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Friends, in last few sessions we have been looking at heterogeneous reactions or in other
words, we have reactions which do not occur in a single phase, but more than one phase
is involved. Now when you have more than one phase involved in a reaction, there are interfaces
that is, the boundaries between these two phases.
As soon as, we have boundaries at the interface or boundary between these two phases, there
is a transfer limitation which comes into picture that is how fast or how slow, the
material gets transferred from one phase to another phase and thus we bring in the transport
phenomena into the domain of chemical reaction engineering and we have spent quite a bit
of time discussing about the kinetics of such processes.
And in this session we will look at few examples or we will try to solve few examples which
relate to these heterogeneous reactions. But, before we do that let us quickly summarize
as to what we have seen in this third module of this course namely on heterogeneous reactions.
So, we started with gas-solid catalytic reactions and discussed extensively on these reactions.
So, we talked about for example, the transport in the catalyst pellet. Firstly, we identify
now that there are several steps, before reaction can take place and these are transfer of reactants
from the bulk to the surface of the catalyst, then from the surface of the catalyst to inside
the catalyst, then adsorption of the reacting species on to the surface chemical transformation
of these active adsorbed species into a product, still adsorbed unto the catalyst surface.
Then, desorption of the product from the catalyst surface it is travel backwards from interior
of the catalyst to the surface of the catalyst and from surface of the catalyst eventually
into the bulk of the catalyst. So, we have two steps which involve external
mass transfer that is, bulk to the surface, two steps which involve internal mass transfer
namely transport within the catalyst pellet and three steps which typically characterize
the chemical reaction that is, taking place by interaction of our reactants and products
with the active surface. So, this transport process will now also play
a crucial role in determining the kinetics of the reaction. Now, how does this transport
process occur? We have bulk diffusion in the or diffusion rather in the bulk phases, but
when it comes to the catalyst pores which can be of micron sizes or even smaller, there
are other factors such as Knudsen diffusion, viscous flow come that is capillary flow also
come into the picture. So, while we can characterize the bulk flow
by the diffusion coefficient, we have to worry about the diffusion coefficient of the Knudsen
regime and the capillary flow and so on. And we have tried to put all this together into
what we call effective diffusivity of a reactant in the catalyst.
The catalyst effective diffusivity is influenced by several factors one of them is of course,
there is multi component diffusion. That is there is no not just binary diffusion species
a diffusing in b, but we have a diffusing in a mixture of b c d and so on and in turn
b diffusing in mixture of a c d and so on. So, that is one modification that comes in.
The second modification, as I said it is a Knudsen diffusion that is if the mean free
path of the gas molecule is large then, there is a possibility that diffusion occurs not
because of molecule to molecule interaction, but because of molecule to wall of the catalyst
pellet interaction and this is referred to as Knudsen diffusion. When can these happen
mean free paths is inversely proportional to the pressure. So, under those conditions
we will have Knudsen diffusion playing an important role.
The third thing is the capillary flow that is, this is equivalent to saying there is
a flow a through a capillary of micron size and there is therefore, a pressure drop between
the inlet of the pore and the exit of the pore and this pressure drop drives the viscous
flow or the convective flow and therefore, we have viscous flow influencing the overall
diffusion process. Apart from this, we also have to worry about
two things that we discussed. What are those, number one is the fact that when we try to
describe this diffusion or diffusion process, we are essentially looking at from the bulk
perspective. So, what is the diffusion of reactant? Let us say a into the catalyst and
this we normally define per unit catalyst area, but we realized that out catalyst is
not an void catalyst that is, the entire area or cross section if you look at the entire
area is not available for flow. What is available for flow is only what are
the void volume or void fraction absolvent void a volume or the fraction to total volume
we call void fraction. So, the diffusivity gets reduced by this factor absolvent that
is one aspect. The second aspect is that we always talk about diffusion in the direction
of the bulk flow. Now, inside the catalyst the path that these
molecules travel is not the same or not in alignment with the bulk flow of the reactant
and product molecules or in other words the path it follows is much more tortuous, because
these pores are not parallel pores from one end of the catalyst pellet to other end. They
can be of any paths with lot of twist and turns and so on or in other words it is a
tortuous path. So, the diffusion coefficients get further reduced by the factor what we
call tortuosity. So, we divide it by the tortuosity factor and this is what determines the effective
diffusivity. We then wrote diffusion reaction mole balances
for these reacting the species. Now moment we say that we have diffusion and reaction
two different processes and within diffusion there is internal diffusion and external diffusion
with each of these process having it is own characteristic rate or time scales. Time scales
are roughly one can in a loose term say inversely related to the rate. If the rate is large
time, scales are small; if rate is small, time scales are large.
So, we now would like to know given this process of diffusion and reaction, how do we characterize,
which are the slow processes, which are the fast processes and so on and for that we defined
what we call the Thiele modulus which rates the time scales for diffusion to time scales
for reaction. So, if diffusion time scales are very large compared to reactions time
scales, the Thiele modulus is has a large value which implies that we are in a diffusion
controlled regime. On the other hand, if the reaction time scales
are very large we are in the kinetic regime. So, this is how we characterize or rate these
two processes and along with this we also rate the reaction rate that we observe. Now,
what is this rating? See, we have diffusion and we have reaction and together determine
what rate we will observe. So, we can consider a scenario, where there was no diffusion and
then do a rating of the observed rate to what would have been the rate, if diffusion was
not there and this rating parameter we call it as effectiveness factor.
We saw how these effectiveness factor and Thiele modulus are related to each other.
For example, when Thiele modulus is very small, effectiveness factor is for most of the reactions
close to unity or rather why most of the reactions all the reactions are close to unity. On the
other hand if the diffusion is very slow; that means, Thiele modulus is very large,
the effectiveness factor varies inversely as Thiele modulus.
We initially had this discussion based on isothermal, first order irreversible reaction
in a catalyst pellet, but then one by one go on extending this concept to a general
kinetics to non isothermal reactions to different geometries of the reactions and saw there
is some kind of uniqueness in relationship between effectiveness factor and Thiele modulus.
What does this Thiele modulus depend on? It depends on the operating conditions such as
concentrations partial pressure for a general kinetics, it depends on the rate constant
of the reaction, it depends on the diffusion coefficient. Then we went on to look at gas-solid
non-catalytic reactions. These gas-solid non-catalytic reaction share several features of gas-solid
catalytic reactions namely the transport aspect is more or less similar.
The major difference between the catalytic and non-catalytic reactions is that the solid
what we are looking at in a non-catalytic reaction is not a static material, but it
is dynamic. That is it is size for example, changes with time and therefore, making gas-solid
non-catalytic reactions in time dependent processes whereas, catalytic reactions we
could easily assume quasi steady state, as far as the diffusion process is concerned
and look at the steady state analysis. What happens, because of this time dependent
characteristic? We actually now have to define how do we calculate the conversion or what
is the time for entire catalyst or entire reaction to be over and this one way of doing
this is to find out for example, if you are looking at case where solid is disappearing,
what is the time required for complete disappearance of the solid such as burning of carbon for
example. In which case our hundred percent conversion
refers to radius of the carbon particle going to zero and then we calculate the time required
and based on this time required this will depend upon the factors whether diffusion
is limiting, reaction is limiting and so on. We also talked about gas-liquid reactions,
once again diffusion becomes or transport phenomena becomes important, but here we focused
more on what happens to the reaction as the kinetics of the reaction changes for example,
when here is a very slow reaction. So, first of all here we assume that the reaction
occurs, because of transfer of material from the bulk gas phase to the interface, then
transfer from the interface on the gas side to interface on the liquid side and one way
common way of dealing with this is to assume Henry’s law some equilibrium relationship
between gas concentration and or gas partial pressure and concentration in the liquid.
Then reaction then transfer from interface in the liquid to the bulk liquid and then
the bulk liquid reaction. So, now depending upon the variation in the
rate of the reaction, we came across various scenarios. For example, if rate is very slow
to the extent that we can neglect reaction in the interface or boundary on the liquid
side that is one scenario. As the reaction rates increase this boundary also starts playing
a crucial role. So, some reaction can take place in the boundary
as well to the other extreme, where reaction is almost instantaneous very fast, and then
the reaction is reduced to actually a plane in which there is a contact between gas and
liquid plane located between the thin boundaries on the liquid side.
Here in gas-liquid reaction one also can talk about rating these diffusion processes and
looking at reaction that we observe with reaction that would have been there, if the diffusion
process was not limiting. But, more convenient and practical way of looking at gas-liquid
reactions is looking at the enhancement factors, because what essentially and this idea comes
from the fact that in absence of reaction there still can be mass transfer and which
would be simple absorption process depending on the saturation concentration of the gas
in the liquid. So, we would have got this rate without any
reaction rate of absorption. Now if you look at gas-liquid reaction even when the reaction
is taking place, we can also still look at the rate of absorption. Our observed rate
is nothing, but the rate at which gas is disappearing from the gas phase gaseous reactant.
So, we can still look at it, but now rate this with what would have been the rate of
these disappearance, if reaction was not there and this rating parameter, we call it as enhancement
factor and now it turns out that because there is a reaction we could have enhancement factors
which by as it is name suggests has magnitudes far greater than one. In fact, you design
your liquid system in a way that enhancement factors are as large as possible.
So, this is how what we basically discussed in our previous modules. So, now let us turn
our attention to few problems in this domain. The first problem says that, we considered
geometries which were spherical cylindrical and so on or flat geometry. But, let us say
that it is in the form of a hollow cylinder and we are supposed to derive the differential
equations that describe the variation of concentration not only as a function of radius, but also
as a function of axial distance. Using suitable dimensionless groups give the
dimensionless form of this differential equation and then ultimately define a suitable expression
for effectiveness factor. Assume that it is a first order reaction, because we want to
keep our discussion simple. But, we have seen enough cases to suggest
that whatever, is applicable for first order reaction with little bit of variation major
ideas being same is applicable for general reactions as well and mass transfer external
mass transfer is also negligible. So, let us try to write this mass balance equation.
So, what is first of all let us start with our geometry. So, we have a cylinder, but
it is hollow. So, let us put the centre point of this as z equal to 0 and let us say half
width half length is h. So, we have z going from minus h to h and again let us put the
centre axis r equal to 0 here and the radius going from a to b.
Let us reverse the order ,it does not make much difference. So, we are getting going
from a to b. Our rate of our reaction is first order, c is the concentration of my reactant,
it is a first order reaction, this two r s are different this is the radial position
and this is the rate of the reaction. So, we keep that in mind.
So, we want to now write a mass balance. So, what we will do is we will consider a small
cross section in the cylinder not in the hollow part, but in the part where the catalyst material
is there and write a mass balance. How do we characterize this small element? The same
way as we normally do us say that this is z and z plus d z along the axial direction
and this is r and r plus d r along the radial direction.
So, the idea here is that we have variation of concentration both along the radial direction
as well as axial direction.
Our mass balance is a mass balance, it is same irrespective of whether, it is a hollow
cylinder, it is a reactor or whatever is there and that is namely accumulation equal to mass,
that is coming in minus mass that is going out plus mass or moles we talk in terms of
mass. So, understood that we are talking in molar concentrations and molar mass basis.
So, this is my mass that is coming in going out we are interested in a steady state balance.
So, this is 0, now let us try to find out what is coming in.
Now, coming in is both from radial direction and also from axial direction. So, let us
talk about radial direction, if my flux is minus D e r dou c by dou r. I am now writing
a small subscript r to say that diffusivity in the radial direction, which could be different
from diffusivity in the axial direction, but that is a detail, but let us call it as minus
D e r at r. So, this is flux. So, has to be multiplied by the appropriate area and what
is that area, this length is d z. So, that area is the surface area d z this
is what is coming in at r; what about z, same thing at given cross section. So, this is
what is coming in multiplied by the area and what is the area, if you look at the z direction
2 pi r d r. So, this is what is coming in. Now, what is going out at r plus d r and at
z plus d z. What is going out will be a similar term. So, minus because in minus out, minus
D e r dou c by dou r at r plus d r area 2 pi r d z. So, that is going out in a radial
direction at r plus d r. What is it that is going out in the axial direction into 2 pi r d r.
So, that is what is going out and what is being generated, what is being generated is
actually this is getting converted. So, that is minus k v into c and what is the area for
this the area for this will be 2 pi r d r d z, rather the volume, because the rate is
per unit volume and the volume of this small cross sectional element is 2 pi r d r d z.
So, this is my general mass balance equations. So, let us repeat once again, we have a hollow
cylinder going from minus h to plus h as it is length and radius from which is the outer
radius to b which is the radius of the inner part and from 0 to b is a hollow portion.
In this to write a mass balance, we selected a small cross section between z and z plus
d z and r and r plus d r. So, this particular small portion and wrote a mass balance for
what is coming in, what is going out plus, what is getting generated.
In writing in and out the idea is flux times the area and for generated, it is rate times
the volume, because the rate is defined in terms of per unit volume. So, that all units
all terms I have same dimensions namely mass.
So, now to this expression, we can divide the entire expression by 2 pi r d r d z and
as we do in deriving any equation take limit as d r goes to 0, d z goes to 0 and if we
do that these mass balance equation will reduce to the following. dou r of d e r dou c dou
r plus r dou z of d z dou c dou z equal to k v into c. This is our mass transfer equation.
So, this is our mole balance, if it is an energy balance involved we could write similar
balance except heat of energy will come and dou t dou r dou t dou z terms will come.
What kind of equation is this, it is a partial differential equation and what is the order
is 2. So, we need two boundary conditions in two variables r and z. So, let us try to
write down these boundary conditions. Now let us start with what is happening at z equal
to 0. What is happening at z equal to 0, z equal
to 0 is a symmetric boundary, symmetric plane. So, concentration gradient at z equal to 0
must be 0 in the axial direction. So, we can write our first boundary condition as dou
c dou z is equal to 0 at z equal to 0 this comes from the symmetry.
Now, what happens at z equal to h? What happens at z equal to h that is at this point what
is happening at that point the gaseous gas is coming out and I do not know what I said
let us say that there is a external mass transfer also. So, that we will have a general module
and then we can we can simplify it. So, what should happen the diffusive flux
at z equal to h must be the flux at which material is going from the boundary of the
catalyst to the bulk, which is k g into c b minus c. This is all evaluated at z equal
to h. What about r equal to a? That means, what is happening at r equal to a. The same
thing the diffusive flux, because material can go out from here as well as from r equal
to b. So, at both these r equal to a and b the diffusive
flux must be same as the bulk the flux from the surface to the bulk of the bulk of the fluid
and therefore, we have these four boundary condition. So, our total mass balance equation plus four
boundary conditions is given in this particular manner. So, let us now try to see how we can
second part of the problem was to convert this into a dimensionless form the equation
should be now dimensionless.
So, now let us define the dimensionless quantities namely rho equal to r by r by a and c star
equal to c by the bulk concentration and another variable zeta which let me use these two and
see how we get that. So, if we use this rho and c star and put it in this mass balance
equation for example, we will get the following. One over rho dou rho of rho dou c star dou
rho plus d e z by d e r a square dou c star by dou z, we have not done anything to z yet,
so k v a square by d r into c star. So, now we can define dimensionless length zeta. So,
that this parameter becomes unity. So, how do we do that we define raise to half
and if we do that then this quantity will become unity and we will get our mass balance
equation in the dimensionless form. No, it should be dou square as immediately we recognized
that this parameter is something which we had done for a first order spherical catalyst
for example, where there was a radius. So, we can define this as a Thiele modulus,
it is a dimensionless quantity phi square c star. So, this is our dimensionless form
of the mass balance equation and then we can similarly convert our boundary conditions
also in the dimensionless form and I will skip that part.
Now, let us come to our effectiveness factor. What is our effectiveness factor? It is the
observed rate so, one over v integral r d v that is the observed rate, which is varying
along the entire volume of the catalyst. So, integral and normalized by the volume that
is volume divided by the rate that would have been , if there was no mass transfer limitation;
that means, my concentration c is same as the bulk concentration.
Now, for a spherical catalyst what is the volume? The volume is difference between the
volumes of the outer cylinder minus the volume of the inner cylinder. So, what is that volume,
that volume therefore, eta is what is that volume of the outer cylinder pi a square into
h or rather 2 h minus pie b square into 2 h. So, into 2 h that is my volume.
Now, what is this d v, remember we have changes in both radial and axial direction. So, from
a to b or b to a from b to a from minus h to plus h k v into c into 2 pi r d r d z.
That is the volume d v and what would have been the rate, if mass transfer was not limiting.
So, that is k v into c b. So, this is my effectiveness factor. So, what needs to be done is to make
this also dimensionless and if you do that we are left with this form of the dimensionless
quantity.
So, that is our definition of our effectiveness factor, which was what our problem was. Let
us look at one more problem. The following data was obtained for a first order irreversible
reaction in a spherical catalyst with a surface concentration of reactant as C 0.0002 moles
per C . So, the data is given. Data is given, so from this we have to find the true rate
constant and effective diffusivity and predict the effectiveness factor and expected rate
of reaction for a cylindrical pellet of dimension 0.5 centimeter by 0.5 centimeters.
So, here the experimental data is given you have to the bulk concentration is given. We
have to do certain calculations to find out what is the true rate and effective diffusivity.
So, this is how we can use our experimental data, because this is our observed rate, which
you can always measure by looking at concentration differences diameter, which is size which
we know. So, based on this can we get information about
the true kinetics of the process and the estimate of diffusion coefficient, which we saw is
an, if modified or effective diffusion coefficient because of this factor. So, let us see how
we go about doing this.
So, what is given to us is the radius diameter; that means, radius versus r observed is given
to us. We are also given that it is a spherical catalyst. So, let us define a Thiele modulus
for a spherical catalyst. Remember k v and d e, k v is the true rate constant, d e is
the diffusion coefficient. Then this is what something, we want to get we do not know what
that value is. So, let us see how we use this information,
this is the knowledge we have, this is the information that is available to us to find
this quantity. So, I know radius which according to what is our phi is some constant C 1 times
radius. What is this C 1? C 1 is nothing, but 1 by 3 square root of k v into d e some
constant. Now what is our observed r observed is eta times r at bulk condition, the bulk
concentration is also given to us. So, the rate at that bulk concentration is
same whether, we do catalyst in one millimeter, ten millimeter or ten meter. So, that is constant,
we can write this as C 2 into eta.
So, now what happens we know that eta versus phi plot is something like this, for a first
order reaction. Now we do not know phi, we do not know eta so we cannot plot this, but
what do we know? We know r observed this quantity. So, if I plot r observed versus I do not know
Thiele modulus, but I know the radius of the catalyst pellet.
So, what do you think should happen? As this two relationship suggest these two axes phi
and r are related to each other through a constant C 1. I do not know what that constant
is, but that is a constant. Similarly, r observed and eta is related to another constant C 2.
Once again, I do not know what the value of that c 2 is, but I know it is a constant.
So, this is nothing, but linear transformation of eta and phi will give you r observed versus
r. So, that also qualitatively will have the same behavior and you want to find out what
is that linear transformation between the two is.
So, now let us rewrite this and see how we can get that, we know for a spherical catalyst
our eta is 3 phi coth phi 3 phi minus 1 divided by 3 phi square. So, for eta I am going to
put from this particular expression r observed by C 2. So, r observed by C 2 and for Thiele
modulus, I am going to put C 1 into r and so if I do that I will get this is equal to 3 into C 1 into r coth 3
C 1 r minus 1 divided by 3 into C 1 into r the whole square.
Now, what do I have? I have r observed versus radius. So, in principal I can do a regression
analysis for r observed versus r with this function, this whole function and get value
of C 1 and C 2. Now, in reality if you want to do a good regression
analysis, you must have lots of data and we have four data points, since then the unknowns
are only two. You would not get the exact estimate, but for class work problem we certainly
can get an estimate.
So, if we now do a regression analysis between R observed versus R, using this function and
then from that regression analysis we will get C 1 and C 2 and that works out to be C
1 is 89.47 and C 2 is 2.56 regression analysis.
Now, what is our C 2? Our C 2 if you go back our C 2 is r of c b. So, C 2 is r of c b,
which according to our module equation is k b into c b. I know my C 2 2.56, I know my
c b which is 2 into 10 rise to minus 4 giving rise to k v equal to 1.28 into 10 rise to
4 hour inverse. What is my C 1? My C 1 is 1 by 3 square root
of k v by d e. I know k v, I know C 1, I can calculate D e which comes out to be 0.177
centimeter square per hour. So, this is how we can get estimates of our true reaction
rate constant and effective diffusivity given the experimental data of rate at different
sizes of the catalyst pellet. The second part of our problem is predict the effectiveness
factor for and the expected rate for cylindrical pellet, this was all for spherical pellet.
Now, how can we predict? If you recall our relationship between eta and Thiele modulus
is more or less same or almost identical. If we define our Thiele modulus properly,
that is for if I define my Thiele modulus as what we had done, just now for spherical particle like
this or this is for my sphere or for my cylinder r by 2 square root of k v by d e, then this
relationship is unique. So, now what can I do? Very simple. I know
r which is given as I think 0.5 by the size is given as 0.5 by 0.5. So, R will be half
of 0.5 0.25, I know k v I just now calculated, I know d e which also I calculated. So, I
know phi c and I know relationship between eta and phi for sphere which I will assume
also applies to cylinder because we had seen this.
So, from that we will get observed rate first of all, we will get phi c which works out
to be about 33.6. So, at this phi c eta of phi at this value is roughly 0.029. You do
not even have to do any calculation, remember this Thiele modulus is much higher than the
value of 3. What is significance? We had seen for Thiele
modulus less than 0.3 eta was close to unity for Thiele modulus greater than 3, eta went
1 over phi. So, moment you know phi you can immediately calculate eta. You do not have
to use any formula, because this magnitude is an order of magnitude higher than phi equal
to 3. So, we have our solution and then r observed
will be eta times r of c b. So, we have all the information and we will get observed rate
which works out to be 0.076. Let us look at another problem, I think this one we had seen
earlier.
Let us look at this problem, based on gas-solid catalytic reaction. What does it say? It says
that two small solids spherical are introduced in a constant environment oven with and are kept
there for one hour. So, this is a piece of information that we
have under these conditions 4 millimeter diameter particle is converted conversion is 0.58.
2 millimeter particle it is 87.5. The reaction in once transformation of a solid reactant
into product reactant, the size is unchanged. An external mass transfer limitation is negligible.
The heterogeneous shrinking unreached core model is adequate. Find the rate controlling
mechanism as diffusion or reaction control and find the time required for complete conversion
of particle in this oven.
This is important, because this is given to us. So, we immediately know what is the time
required for complete conversion. If we take all the three phenomena that time and this
we had seen or you can see these expression into conversion x contribution from external
mass transfer plus, what are various terms here r is the radius the initial radius of
the particle, rho s is the density of the density of the particle, d 1 is the diffusion
coefficient. So, this is corresponds to contribution due
to internal diffusion plus r rho s, rho s is the solid density c b is the bulk concentration
x is our conversion, this is if reaction is controlling. So, we have the time required
as a function of conversion for a general scenario, when all external mass transfer,
internal mass transfer and reaction dominate or play an important role.
Now external mass transfer is negligible here. So, this contribution is 0. So, now we have
to examine whether for the data that is given to us. What is that data? we have been given
two sizes particles of two sizes and in one hour; that means, in the time one hour what
is the conversion for this two different sizes particles. So, we have to find out whether
internal transfer controls or reaction controls. Now, how do we do that. So, let us assume
that internal controls or internal diffusion controls, then we will get time as that means,
this term is negligible only first term is applicable.
So, r square rho s by 6 D 1 e into c b into 1 minus 3 into 1 minus x two-third plus 2
into 1 minus x. So, this is if internal diffusion controls that means, first and third term
are negligible. Then, based on information that is given to us. What is given to us is
r conversion. So, two different particles and conversion. So, let us say r 1 x 2 is
known to us, r 2 x 2 is known to us and time is same 1 hour.
So, knowing time t equal to 1 hour for r 1 and I will keep these terms rho s, because
that is constant, I do not know what that value is should be 1 minus. Let us say this
is conversion is
this must be and if internal conversion was the same, this must be also r 2 into corresponding
same term into one minus, whatever was the conversion there, that is 875 rests to 2 by
3.
So, if this was correct then, the ratio of the two should be equal, because that is the
same one hour. Now it turns out that the ratio of r 1 to r 2. So, what must happen r 1 to
r 2 plus 2 into 1 minus 0.875 as it is complete, r 1 by r 2 must be this factor over here So,
I will just show it like this divided by this factor over here.
The actual r 1 to r 2 is 1 by 4 r 1 square, actually this square r 1 square by, but this
is not 1 by 4 not equal to 1 by 4 that means, diffusion is not the correct. Now what will
happen, if I assume reaction is reaction is controlling then I get this.
So, I can do the same exercise r 1 r 2 and it turns out that ratio is correctly matches
that means, the reaction control is the correct controlling regime. So, with this few examples
we will stop for this session and continue to our discussion on the last module, in this
course namely reactor design thank you.