 Friends, it is a good time to summarize what we have learnt in the last couple of lectures. So we are looking at the fluid solid heterogeneous, non-catalytic heterogeneous reactions. And then we identified two possible cases, two possible modes. One is the situation where the particle which is actually participating in the reaction, it does not shrink with as the reaction progresses. And then another situation where the particle which participates in the reaction actually shrinks in size as the reaction progresses. And then we looked at two possible models, progressive conversion model and the shrinking core model. And it was observed that the shrinking core model is more common in most of the situations. And so we developed a model for the shrinking core model for both the particle size, unchanging particle size and also for those particles which were shrinking in size. And that for situations where the overall reaction, overall conversion is actually controlled by different rate controlling steps. For example, the diffusion of gas phase from the bulk through the gas film to the surface of the core or the diffusion of the species through the ash layer or it is the reaction controlling. So the expressions that we get for the time that is taken for the unreacted core to reach a certain size which is directly correlated with the conversion of the solid species can be summarized as below. So for spherical particles, so it can be summarized in this table here. So for a spherical particle suppose we take the constant size and varying or changing size. These are two possible modes. Here changing size refers to the special case of no ash layer present. As noted in the earlier lecture, the size can change even if the ash layer is present if the densities of solid product and the reactant are different. Now suppose if it is gas film diffusion control, if the overall reaction is controlled by the film diffusion control then we found that T by tau where T is the time taken for the unreacted core to reach a certain radius because of the heterogeneous reaction that should be equal to the conversion of the solid itself which is directly related to the amount of solid that is reacted. And then in changing size the expression for the same would be T by tau is 1-1-xp to the power of 2 by 3. Here tau is the time taken for the complete conversion that is all of the solids which is present in the core has gone for complete reaction. Now next possibility is ash layer control, ash layer diffusion control. So if ash layer diffusion is actually controlling the overall conversion in that case T by tau is given by 1-3 into 1-xp to the power of 2 by 3 plus 2 into 1-xp. That is the expression for T by tau and of course in changing size the ash layer does not exist and therefore it does not offer any resistance to the overall conversion. Then the third case is where the reaction is controlling the overall conversion, it is the reaction control then T by tau is given by 1-1-xb to the power of 1 by 3 and here T by tau is given by 1-1-xb to the power of 1 by 3. In both cases the time taken for the unreacted core to reach a certain radius depends on the conversion in exactly the same way because the presence or absence of the film or the ash layer does not contribute to the overall conversion if it is in the reaction control regime. Now all this we looked for spherical particles. What happens here? What happens if it is other geometries? For example if there is plating of metals then normally it is not plated on a spherical particle it is plated on other kinds of geometries, it could be a cylindrical geometry or it could be a flat plate geometry or it could be any other curvature. So let us look at two specific other geometries which are fairly common. One is the flat plate, suppose if it is a flat plate and where the solid is now present in the flat plate and there is a fluid which is now coming on to the surface of the flat plate and the heterogeneous reaction is occurring at the surface of the flat plate. So in that case if we consider the constant particle size, if we consider the case of constant particle size then if the for a flat plate one can actually define one can define conversion as 1-1 by L where L is essentially the half thickness, L is the half thickness of the plate that is actually being used for used as a solid reactant and if it is a film diffusion controlling, if it is a film diffusion controlling regime, in that case T by tau is given by xp where tau is given by tau is the time taken for the total conversion that is for all the solids to undergo reaction that will be given by density of the particle divided into the half thickness length scale divided by mass transport coefficient into concentration of the species in the gas phase. If it is ash layer diffusion control, then T by tau is essentially given by xp square where tau is given by rho B L square divided by 2 diffusivity into concentration of the species in the gas phase. If it is reaction controlled, then T by tau is simply given by xb and tau is given by rho B L divided by k double prime which is the intrinsic rate constant multiplied by CAg. K double prime is the intrinsic specific rate and CAg is the corresponding gas phase concentration. Now similar expressions have been obtained for cylinder geometry and they are as follows in this table. So for cylinder, for a cylindrical geometry the xp conversion of the solid is defined as 1 minus R by R0T whole square. Now here R0 is the initial radius of the core, R0 is the initial radius, initial radius of the cylinder and R is the instantaneous radius, instantaneous radius of the cylinder. So with this definition, if it is gas phase diffusion control, then T by tau is essentially given as xb and tau is equal to rho B into R0 which is the initial radius divided by 2 times kg into CAg. So that is the time taken by the core for complete conversion and if it is ash layer diffusion control then T by tau is given by xb plus 1 minus xb multiplied by the natural logarithm of 1 minus xb where tau is tau B, tau is the time for total conversion that is given by rho B into R0 square divided by 4 de into CAg. So that is the concentration of the species at the gas phase and if it is reaction control, if it is reaction control then T by tau is essentially given by 1 minus 1 minus xb to the power of half and tau is equal to rho B into R0 divided by k double prime that is the specific reaction rate multiplied by CAg. So this sort of summarizes the various time taken for the core to reach a certain radius if the overall reaction is controlled by different resistances which is actually available in the system. So all these are for basically constant size system and similar expressions can actually be worked out for varying size as well. We so far looked at the cases where the overall reaction is actually controlled by the resistance of one of these three resistances which are available that is either the diffusion through the gas film or the diffusion through the ash layer in the case of constant particle size or if it is a reaction controlled. So in reality this is not the situation, in reality what happens is that all three resistances actually contribute. So as reaction proceeds and in fact that is true because as reaction proceeds, as reaction proceeds even in the constant size case, even the constant size case what is observed is that the of course the unreacted core radius is going to decrease, unreacted core radius decreases as the reaction proceeds and therefore as a result the relative importance of various resistances is going to change. So that is very important because there are three resistances as the size of the unreacted core changes then the relative importance of these different resistances towards their effect on the overall conversion also is going to change. So which means that not one, so no one resistance controls overall conversion all times, all time t. So therefore it is important to consider simultaneous action of all the resistances, so it is important to consider simultaneous action of all three resistances. So if you want to incorporate all three resistances then the model for the radius of the unreacted core as a function of time that the differential equation which basically tells what is the rate of change, relates the rate of change of the radius of unreacted core with respect to time to all the other properties and concentrations of the system can actually be rewritten using a combination of all the resistances and the rewritten model will be dr by dt that will be equal to minus CAg divided by rho b, CAg is the gas phase concentration rho b is the density of the solid that is used divided by r square by r naught square into kg plus r naught minus r divided by the diffusivity into r by r naught plus 1 by k double prime that is the specific reaction rate. Now here it is assumed that the volume fraction of the solid in the unreacted core is approximately equal to 1 and so there are three terms here which correspond to three different resistances. So the first one here this corresponds to the gas film, this corresponds to the gas film resistance and this corresponds to the ash layer, this corresponds to the diffusion in ash layer and the last term corresponds to the reaction resistance, resistance due to the heterogeneous reaction that is occurring in these on the surface of the unreacted core. So now suppose this is for a constant size system, so if the particle size does not change, this is for the constant size system. Suppose if the same thing can be written for a shrinking size type of particles where there is no ash layer, so if there is no ash layer then the resistance due to ash layer does not exist. So simply we can obtain the expression by removing the resistance due to the ash layer which is present in this expression here. So for shrinking particles there is no ash layer, so therefore we can write this expression as dr by dt is minus Cag by rho B divided by 1 by kg plus 1 by k double prime where k double prime is the specific rate and kg is the corresponding mass transport coefficient. So now let us take an example problem and see how we can predict the dissolution of a particle of a certain type which is undergoing such kind of a fluid solid non-catalytic heterogeneous reaction. An excellent example of that is the drugs which are actually being administered. So the solid drugs which are administered for curing a certain disease, when they go into the body they have to dissolve and it is important to understand how much time does it take for the drug which is a solid particle to dissolve and that the dosage of the drug actually strongly depends on the time taken for complete dissolution. So therefore it is important to model this from the pharmacokinetic standpoint of you. So let us look at this problem. So dissolution of solid particles, suppose if we assume that there are all these drugs which are given inside, they are all mono dispersed particles that is all the particles that are actually in the drug are of same size. This is not true always but let us assume to start with that all particles are actually same size that is it is a mono dispersed system and then so this is, so we need to, the objective is to find out what is the dissolution time which actually if we place an important role in the pharmacokinetics. Now suppose if there is species A which is basically in the fluid stream that reacts with certain solid which is essentially the drug particle and that leads to formation of products. Now if we assume that the fluid A actually reacts with a core which contains the solid material and if we assume that it is first order with respect to the fluid, the reaction rate is first order with respect to fluid and zero order with respect to the solids which is present. Then the question is what is the dissolution time? So we need to estimate the objective of the problem is to find the dissolution time, the time that is taken for dissolution of these particles. So now we can observe, we can actually observe that the rate of mass transport, the rate of mass transfer to the surface is actually equal to the rate of surface reaction. Why is this because the particles are expected to as the property of the drugs are always such that the particles are expected to dissolve and so the particles are, this is actually a process where the particles are actually shrinking in size. So there is no ash layer which is actually present. So therefore the rate of mass transfer to the surface should be equal to the rate at which it is being consumed for the surface reaction. So therefore we can write the WAR if that is the flux at which the species is actually traveling from the bulk gas phase to the surface of the solid that is equal to kg into Ca minus CaS where Ca is the bulk concentration and Ca is the surface concentration. And that should be equal to the reaction rate Ra double prime and that is equal to if Kr is the specific reaction constant multiplied by CaS. So that is the surface concentration and this is the net reaction rate. So this is the reaction rate and this is the mass transport. This corresponds to the mass transport. So from here we can actually eliminate CaS. So note that CaS is a quantity which cannot be actually measured experimentally but Ca is the bulk concentration and that can actually be measured. So we need to eliminate CaS which is not a measurable property, measurable quantity. So kg into Ca divided by kg plus Kr. So plugging in this back into the expression for WAR we find that double prime that is equal to Kr into kg Ca divided by Kr and that is equal to Ca by 1 by kg plus 1 by Kr. So therefore if we know the mass transport coefficient and the reaction in specific reaction constant reaction rate then we will be able to actually calculate the required quantities just the dissolution time. So therefore now there is a need to find out what is this mass transport coefficient. So we can use correlations in order to find the mass transport coefficient kg. So suppose if we assume that the particles are small. So under small dp and if there is no shear stress at the, so if there is no shear stress at the boundary of the fluid and the solid then we can actually using the Frosling correlation we can find out that from Frosling correlation we can find out that Sherwood number which is equal to the mass transport coefficient into diameter of the particle divided by the corresponding fuselage that is approximately equal to 2. So from here we can find out that the mass transport coefficient is given by 2 times diffusivity divided by the corresponding particle diameter of the particle which is actually being dissolved. So plugging in the back into the expression for flux we find that WAR equal to Kr into C A 1 plus Kr by kg and that is equal to Kr into C A divided by 1 plus Kr into dp divided by 2 times de and with this can be rewritten as Kr into C A divided by 1 plus dp by d star. So where d star is nothing but the ratio of 2 times de into Kr so that is what is d star. So d star the meaning of d star is that it basically tells you the diameter at which the mass transport and the reaction rates are actually equal. So if we look at what is the expression for d star, d star is 2 times de divided by Kr and that is essentially the diameter at which the mass transport rate and the reaction rate are equal. So therefore if the diameter of the particle if this is larger than d star then it can be expected that it is mass transport controlling while if the particle diameter is less than d star then it is actually expected to be a reaction controlling scheme. The overall conversion is expected to be reaction controlling. So now in order to find the radius of the particle as a function of time we can now write a mole balance on the solid particle and the mole balance goes as here, mole balance on solid particle. So where whatever rate at which things are coming inside minus whatever rate at which the particles are leaving out plus generation should be equal to accumulation that is the balance. Now nothing is coming inside because it is the solid which is actually participating so there is no flow so nothing is coming inside and nothing is actually leaving but some of it is actually being reacted so that is a counter to the generation term. So if rbs is the rate at which the solid is being consumed at the surface of the particle multiplied by pi into dp square that is the surface area of the solid particle and that should be equal to d by dt into density into volume of the spherical particle. So suppose if we assume equimolar counter diffusion, suppose if we assume equimolar counter diffusion then the minus rAs double prime that is the rate at which the species A is actually being consumed because of the heterogeneous reaction that should be equal to minus rbs the rate at which the solid is being consumed this is because of the equimolar counter diffusion and the stoichiometry which is associated with the particular reaction. So now this equation can further be simplified as rho into 3 pi by 6 into dp square into that is equal to rAs into pi into dp square. So this can be simplified as d by dt of dp that is equal to minus 2 minus rAs double prime divided by rho. So that is the expression for the rate of change of the diameter of the drug particle as a function of time and how is it related to the surface reaction rate and the density of the particle. So now this can actually be we know the what is the rate so that can be written as minus 2 into kr into Ca divided by rho into 1 by 1 plus dp by d star and this can be written as minus alpha divided by 1 plus dp by d star. So that is the expression for rate of change of the diameter of the particle as a function of time. So if you assume that this whole thing is a constant like alpha so we can rewrite this expression as d by dt of dt equal to minus alpha by 1 plus dp by d star where alpha captures the rate of the reaction divided by the density of the corresponding particle. So now we can integrate this with the following initial condition that at time t equal to 0 the initial size of the particle was dpo and based on this initial condition the equation can be integrated and integrated expression would be dpo minus dp plus 1 by d star 1 by 2 d star into dpo square minus dp square that should be equal to alpha t. So this provides a relationship between the properties of the system and the diameter of the particle and time. Now in order to obtain a complete conversion for this problem for attaining achieving complete conversion that is for complete dissolution of the drug so for complete dissolution of the drug the dp should be equal to 0 that is all the particles are actually completely consumed and so on and the time tau which is the time taken for complete conversion is given by 1 by alpha into dpo plus dpo by 2 d star square. So that is the relationship between the time taken for complete conversion and the diameter of the particle and other properties of the system. Now this is the expression this is for a mono dispersed particles but normally the drugs which are actually administrator they are all poly dispersed that is particles are actually of different sizes. There is a population of particles and each of these particles can actually be of different sizes and all of them are actually simultaneously undergoing dissolution leading to the shrinking of the particles. Now the only issue here is that different particles will actually do different things because the sizes different although the reaction is same their dissolution rates are expected to be different because their sizes are completely different. So let us look at dissolution of let us see how to capture this behavior dissolution of poly dispersed particles. So what is poly dispersed particles? Suppose you have collection of particles which are of different sizes for example the this big one could be of size d1, it could be d2, maybe it could be d3 and this could be d4 and so on and so forth. So there will be a collection of particles and each of these are different in size and so there are particles of different sizes. Now not just that the initial stage that the particles will be of different size there will be a distribution of sizes of particles all through the time and the dissolution is actually occurring. So therefore clearly there is a distribution there is a size distribution and the question is to find out what is the dissolution time, the objective is to find the dissolution time and in order to find the dissolution time for particles who are actually placed in a certain distribution one needs to actually follow the distribution dynamics. So the distribution dynamics have to be followed. So let us look at what is the distribution. So suppose if I look at the plane of number of particles versus the fraction of the number of particles that are actually of a particular size. So then we expect that it is a certain histogram where the location between let us say dp and dp plus delta dp which is a small increase in the particle, small difference in the particle size. So in that case the area and under the curve in this location it signifies the number of particles which are actually present in the system whose diameter is between these two numbers. So therefore f dp multiplied by delta dp that is basically the number of particles between dp and dp plus delta dp. So that is the number of particles and moreover if we integrate this expression from 0 to infinity f dp into dp into the differential of diameter of the particle that should be equal to n naught which is the total that is the total initial particles which are actually present. Now if we assume that it follows a log normal distribution let us assume that the initial distribution of the particles they follow a log normal distribution. Let us assume that they follow log normal distribution. So now the fraction of the species, fraction of the particles whose size is dp at time t equal to 0 and that normalized by n naught is actually given by 1 by dp square root of 2 pi into ln of sigma 2 multiplied by exponential of minus ln of dp by dg and square of that divided by 2 times ln sigma 2 square. So that is the expression for the log normal distribution where dg and sigma 2 are essentially the distribution parameters these are the distribution parameters. So now the question is in order to find out what is the dissolution time of the particles which are actually in this polydispers system is to actually find out what is the how the distribution itself changes instead of monitoring every drug particle it is better to monitor simply the population itself. So the question is how does the how does distribution change? So that is what we are going to look at. Now suppose if we look at the distribution at any time and if it is some general curve like this now we can write a simple balance on the population of these drug particles and suppose if we take a small element between dp and dp plus delta dp. So therefore the thickness of this element is delta dp if we assume the thickness is delta dp and by definition of this histogram the integral of 0 to infinity f of dp, t into ddp should be equal to n0 that should be the total number of particles that were actually initially present and because there are no new particles are added it is only the particles actually dissolve and disappear. So therefore the integral under the curve up to infinity if we assume that the particles are actually of some finite size at all times then that should be equal to the total number of particles itself. Now if that is not the case then this should be a function of time. So the number of particles will change because some of these particles will dissolve and disappear and when that happens then this integral between 0 to infinity should actually be a function of time. Now one other quantity which is required in order to model the system is the growth rate. Suppose rdp is the growth rate of every particle and of course it depends upon the size of the particle itself. So that is the growth rate, growth rate of particle whose diameter is dp then one can actually write a population balance equation in order to capture the dynamics of the whole distribution itself as the reaction actually proceeds. So the population balance equation can be written as the balance between the number of particles which are actually growing and reaching the diameter whose sizes between dp and delta dp and also we need to account for the particles which are already present in this small element dp and delta dp and they grow and they actually become bigger than dp plus delta dp. So which means they leave this small element delta dp and there is no addition of new particles because once the drug is being fed it is just being dissolved and so there is no other mechanism by which the particles are actually being added into that small element delta dp and then the other term is the accumulation term. So therefore putting them all together it will be number of particles growing into region between dp and dp plus delta dp minus the number of particles growing out of region between dp and dp plus delta dp that should be equal to the accumulation of particles in delta dp in that small element. Now suppose if r is the growth rate which is what we have defined a short while ago then the number of particles growing into the region between dp and dp plus delta dp is actually given by r into dp. So that is the growth rate when the particle is right whose size is exactly dp and then that multiplied by the corresponding fraction will till will provide a clue as to what is the number of particles are actually growing into the region in this small interval of diameter and that evaluated dp minus delta dp and that should be equal to d by dt so that is the accumulation of the particles in delta dp equal to f of dp comma t multiplied by delta dp. So that is the rate of change that is the rate of which the particles are actually being accumulated in that small element. So now rearranging this equation and setting delta dp equal to 0 so set delta dp to be 0 then can be written as minus d by d of the particle diameter into f of dp comma t that should be equal to dou f by dou comma t. So that is the expression for that is the population balance. So now what do we do with this population balance? So we can solve this equation in order to find the radius of the particle in order to find the how the distribution the radius distribution actually changes. So now this can actually be rewritten as df by dt plus r into df by dp plus f into dr by dp that is equal to 0. So now where rdp is essentially the growth rate of the particle of size dp and if suppose in the earlier case if the particles were mono dispersed we actually found out what is the growth rate. So therefore for every particle of a certain size that expression can be used as a growth rate of the particle of that particular size. So therefore from here it will be ddp by dt that should be equal to minus alpha by 1 plus dp by d star. So that is the rate at which the mono dispersed particles are actually growing. So plugging that into the expression plugging that into the population balance we will find that so this is for the mono dispersed case. So plugging that into the expression we will find that the rate of rdp that is equal to minus alpha by 1 plus dp by d star. And so the population balance will be df by dt plus minus alpha by 1 plus dp by d star into df by dp plus f into alpha by d star into 1 by 1 plus dp by d star the whole square that is equal to 0. So that is the expression for the that is the population balance that actually captures how the dynamics of this distribution actually changes with time, how the distribution changes with time. So now if we introduce a few dimensionless quantities which is actually useful in solving this problem so psi equal to f into d star divided by n naught where f is the distribution of the particles based on the size and d star is the ratio of the diameter when the mass transport rate and the reaction rate are equal to each other and n naught is the initial total number of particles. Then epsilon is defined as 1 plus dp by d star and theta is defined as alpha into t by d star. So now if we introduce these dimensionless quantities we can rewrite the population balance as d psi by d theta that is minus 1 by epsilon into d psi by d epsilon minus that is equal to minus psi by epsilon square. So that is the expression. So this equation is actually of a very, very familiar differential equation form which is of the form p of x, y into dz by dx plus q of x, y into dz by dy that is equal to r of x, y, z. So from here we can see that if p of x, y is nothing but 1 and q of x, y is like 1 by epsilon and the r of x, y is like psi by minus epsilon square. So the way to solve this equation is basically to use the method of characteristics is to use the method of characteristics where this problem is posed in a slightly different way. So dx by p sorry d theta by 1 is equal to d epsilon by minus 1 by epsilon and that should be equal to d psi by minus psi by epsilon square. So that is the way in which the population balance equation which is actually written here can actually be posed in terms of the method of characteristics. So now the first two terms in the in the repose problem basically it looks like d theta by 1 is equal to minus epsilon into d epsilon. So from here we can find out that epsilon square plus 2 theta is equal to a constant c 1 and that should be of the form of the form h of epsilon square plus 2 theta. So that should be the functional form functional dependence of this particular expression and then next let us look at the other case d epsilon by 1 by epsilon that should be equal to d psi by psi by epsilon square. So this can actually be rewritten as d epsilon by epsilon is equal to d psi by psi which can be solved to obtain ln epsilon equal to ln psi and that is equal to some constant k. So that is a constant. So from here we can find that psi by epsilon is nothing but some constant c 2 and that can be equal to h of epsilon square plus 2 theta. So that is the functional dependence. So next we can actually see how to take this forward and find the distribution profiles. So suppose if we can rewrite the relationship between the psi and epsilon as psi equal to square root of epsilon square into h of epsilon square plus 2 theta. So in order to obtain the final distribution which is actually present in the definition of psi. So psi is almost like a non-dimensional distribution of the various size of the particles which are of different sizes. So if you look at the initial distribution which is actually given by log normal distribution. So the log normal distribution is as it goes here root 2 pi into ln sigma 2 into exponential of minus ln dp by d star dp by dg the whole square divided by 2 ln sigma 2 and square of that. So that is the initial distribution. Now if you know the initial distributions then we can actually introduce the transform variables. So basically we introduce epsilon psi and theta into this initial distribution and if we assume that epsilon equal to 1 plus dp by d star. So this comes from the non-dimensional form of epsilon. And from here we can get that dp is equal to d star into epsilon minus 1. Actually this is the non-dimensional form of epsilon and from here we can find out the relationship between dp and the other parameters. So if you assume that dr, if you assume a new variable dr is equal to dg divided by d star where dg is some parameter of the distribution and d star is the diameter at which the mass transfer rate and the reaction rates are equal. Then we can find out that psi is equal to some f of epsilon, theta divided by n0 into d star that should be equal to square root of epsilon square into h of epsilon square plus 2 theta. So that is the functional form of this variable psi. So now if you want to find out what is the final distribution we need to find this expression f of epsilon, theta. This expression needs to be found in order to find out what is the instantaneous distribution of the sizes. So how can we do this? So it can be done by actually using the distribution at the initial state where the reaction has not started then f of epsilon, theta. So this can be obtained by simply replacing epsilon square with plus 2 theta in f of epsilon, 0. So note that this is the initial distribution and the final distribution can simply be obtained by replacing epsilon square in this initial distribution with epsilon square plus 2 theta. So that serves as the solution methodology. So now we can rewrite this as psi of epsilon, theta. So that is equal to f of epsilon, theta. So we can introduce, we can now use the initial distribution in order to find out what is the value of psi, what is the expression for psi as a function of epsilon, theta and that would be equal to epsilon divided by square root of 2 pi into ln sigma 2 into 1 by square root of epsilon square plus 2 theta into 1 by square root of epsilon square plus 2 theta minus 1 multiplied by exponential of minus ln of square root of epsilon square plus 2 theta minus 1 by dr and square of that divided by 2 into ln sigma 2 to the power of square of that. So that is the distribution at any time in the non-dimensional form. So once we know this distribution, we can look at the distribution profiles as a function of time. So suppose if this is dp in the particle diameter and this is f of dp, time then if suppose this is the initial distribution, so that is time t equal to 0, that is the initial distribution, then as time progresses all the particles are actually undergoing the dissolution because of the heterogeneous reaction and so as time goes by the distribution changes and so this is t1 which is greater than 0 and then as time further goes by more dissolution will happen and so this will be t2 which will be greater than t1. Then further time elapses and then the distribution will be this will be t3 which is greater than t2 and then eventually when the conversion is going to be almost complete, then the distribution will look like this where this is t4 which is greater than t3. So that is the kind of distribution that one can actually obtain as a function of time for this polydispersed particles which are undergoing dissolution. So let us summarize what we have learnt in the last 3 lectures. So we have looked at the fluid solid non-catalytic heterogeneous reactions and this can occur through 2 modes. One is where the particle size does not change and other one is the case where the particle size actually changes as the reaction progresses and there are 3 different possible resistances which are actually existing in this kind of a system. One is the gas film resistance for diffusion resistance and the other one is the ash layer diffusion resistance and the third one is the reaction controlling resistance offered because of the surface reaction. So these 3 under these 3 regimes the time that is taken for the time that is taken by the particle to reach a certain radius has been modeled and calculated and that for different geometries have been looked into for both the shrinking or the changing size particle case and for the constant size particle case. And then we looked at an example of dissolution of monodispersed particles and extended for a polydispersed system. Thank you.