 Good morning. So, we will continue our discussion on effectiveness factor. Now, today we like to wind up this particular topic. We have already looked at what is effectiveness factor. It is something that tells you the effect of pore diffusion resistance on the overall rate of the reaction. It is related to a dimensionless number called as Thiele modulus. There is an inverse relationship between eta and phi. Eta is effectiveness factor. Phi is dimensionless number that is Thiele modulus. And effectiveness factor normally varies between 0 and 1, but there are some unusual or some exceptional cases where effectiveness factor can be greater than 1. So, what is effectiveness factor? In general, effectiveness factor is the rate of the reaction observed rate of the reaction which may be hampered by the pore diffusion divided by the rate of the reaction that is calculated at the external surface as if there is no inter particle pore diffusion resistance. So, most of the times it is less than 1 because there is some resistance. If at all there is no resistance, it is 1, but which are the situations where we come across case where like the rate inside a particle becomes greater than overall rate. Of course, I have raised over the entire particle radius. The rate is greater than the rate that is calculated based on external conditions. So, I told you towards the end of the last lecture that eta can be greater than 1 when it happens mostly in the case of hexothermic reactions, when in the particle the temperature inside and temperature outside. So, T i is greater than T s. When does that happen? It happens when you have exothermic reaction, the heat is liberated because the reaction takes place inside a particle, pore diffusion resistance is significant. So, it does not take place only at external surface. It goes inside reactant and then the reaction takes place and the heat is liberated. Now, this is another condition that once the heat is liberated temperature will rise, but then if the heat gets dissipated fast from the interior part of the catalyst particle to the external part. Then this dissipation rate will allow the temperature to go down or come down to a particular level and probably it will be equal to the external temperature. But, if the dissipation rate is relatively low in that case, the temperature inside is going to go up and when will that happen? This is all at steady state. When will that happen? It will happen when the conductivity, the thermal conductivity of the particle. Now, this again effective thermal conductivity like effective diffusivity because the porous structure. So, the thermal conductivity of the particle is less than the or rather it is relatively low rather. So, that a heat dissipation is not that fast. So, temperature inside a particle T i is greater than the temperature outside a particle. Now, the rate of the reaction as we normally say rate of the reaction is typically at of course, when the particle or molecule is close to the catalytic side, the rate of reaction is k into C A raise to n where I did for a then minus. Now, this k is a 0 exponential minus e by r T C A raise to n. Now, if you look at this is a so a or a 0 this is frequency factor. Now, the two parameters which affect the rate, the temperature and concentration normally in isothermal case temperature is same throughout and same as the external surface temperature is only the concentration that changes in a normal situation or whatever we have discussed before this that is for isothermal case the concentration inside a particle is less than concentration in the external surface. So, this reduces the rate inside a particle. But now, look at this expression not just C A, but T is also changing. Now, this T inside is greater than the T at external surface. So, these two parameters C A and temperature they will affect the rate, but relationship are opposite that is increase in temperature would cause increase in rate. But there is a reduction in concentration because of the pore diffusion effects that will reduce the rate. So, now the overall effect of these two changes that is temperature and concentration on the rate is going to decide whether the internal rate or rate inside a particle overall rate average or other integrated over the entire particle that rate is going to be higher than calculated that at the external surface. So, there are situations where it is likely that temperature would go up such that it will compensate for the effect of concentration and the rate inside a particle would be greater than that calculated at the external surface in that case effectiveness factor becomes 1 or greater than 1. So, let us look at the relationships. So, let us plot a graph of phi say for a first order reaction phi 1 and eta again you have 1 here then say 0.001 is again a log log plot 0.1, 1, 10 and 100. Now, this is 1 and you know the famous or other whatever relationship that we have already looked at is this. So, initially or other low values of phi it is constant and close to 1 and then it goes on decreasing this is what we have seen and that is for isothermal effect that mean there are no heat effects. So, let us define a parameter let us define a parameter called as beta which is delta T max let me tell you the meaning of this delta T max divided by T s. Now, what is delta T max delta T max is a maximum possible temperature rise that can happen inside a particle and T s is the external surface temperature now this particular delta T max can be calculated based on the heat generated. So, heat generated would depend on the heat of reaction. So, delta H r of the reaction this is a heat generated into D e C a s divided by K T again delta T max. Again I will tell you meaning of each and every term is heat of reaction you know this is diffusivity this is external surface concentration and this is thermal conductivity this is thermal conductivity. Look at units this is a dimensionless number the dimensionless number and this tells you so the numerator tells you about heat generated some proportional to heat generated maximum amount of heat generated this is exothermic reaction. We will see what will happen in case of endothermic reaction later but right now we talk about rising temperature. So, and the rise will depend on the thermal conductivity the gradient. So, this particular thing is delta max and this is the T s. So, for this particular case beta is 0 because we are saying that there is no heat effect we are saying that there is no heat effect. So, what we have seen so far is for beta equal to 0. Now, this is another parameter that needs to be defined of course we can look at a elaborate derivation of all this, but not going to spend much time here this is for your information, but if you are interested you can write energy balance also inside a particle like what we have done in the case of component balance we write energy balance also. So, now you have two equations component balance energy balance play with those equations. So, all those equations and you will get a value of eta. Now, again it has to be done numerically because now we do not have a straight forward relationship for eta like what we had before what was it eta is equal to 3 divided by phi 1 square in the in the bracket phi 1 cottage phi minus 1 phi 1 rather. So, phi why phi 1 because it is a first order reaction. So, we have derived a relationship of first order reaction spherical particle, but then that was for isothermal case. Now, for non-isothermal case we will have energy balance also coming into picture and then the solution becomes slightly complicated getting analytical expression becomes difficult, but you can do it numerically. So, you can do it numerically. Now, this is for beta equal to 0. Now, if I increase the value of beta you will start seeing a train like this slightly higher value of beta then this then this then this and of course, you can further do this. What is happening here? The value of beta is increasing. Now, in order to do these calculations you have to do these calculations I will have to define some other constants also and there is another number called as gamma which is nothing, but E by R T S or delta E by R T S. Now, what is delta E? It is the activation energy. So, it tells you about the sensitivity of your reaction rate constant or reaction rate intrinsic reaction rate towards the temperature that is also important because if the reaction is not sensitive to temperature at all then there is a problem. Then even if temperature rises inside a particle it does not have any impact on the rate. So, it is as good as no heat liberated as far as reaction is concerned because reaction is not going to be affected by the temperature rise. So, the effect of temperature rise is who decides that is the activation energy. So, we need to define the activation energy. So, there is another parameter. So, what the graph that I have made here for different values of beta it is at a particular value of gamma. So, this graph is at certain value of gamma right and for certain value of gamma I go on increasing the beta value one of these can change. So, that I get different value of beta right. So, in general beta represents a maximum temperature rise and that depends on many factors and the most important one is the heat of reaction and of course, the thermal conductivity. I said like if your thermal conductivity is small dissipation rate would not be large and that is why the temperature rise will be large. So, that is why K T appears in the denominator look at all this. So, if you try and read these equations try and interpret them try and see the effect of different variables parameters right. So, beta shows the effect of heat of reaction. Now, it is a very peculiar trend that you are seeing. So, as the value of beta increases as the maximum temperature rise increases as the internal rate increases compared to that of the external rate or rate at calculated the external conditions then the value of eta increases. So, this is one and now you can see they are all values here greater than one and that happens when the heat effects are significant at large values of beta. You will get a similar such graph for different values of gamma. So, there are different parameters which would affect the value of eta and for non-osothermal case we have two independent parameters defined it is beta and gamma. So, eta can be greater than 1. Now, the here I can see a very unusual trend or unusual sense like you can imagine what will happen if you have a value of 5 say 0.08 or whatever let me use another color. What am I seeing here? I am seeing that there is a possibility for the given value of 5. There is a possibility that you have one value of eta here one here and one here what does it mean? This is a solution that is obtained by solving steady state equations for material or component balance and energy balance. So, I am solving these equations together and I am getting a value of eta for different values of 5. So, suppose the value of beta is very large then I am seeing this I am seeing this trend I have got this trend where you have a typical S type curve here S type curve here. Now, this S type curve rather gives rise to three different solutions at the same value of 5. How do I interpret this? How do I interpret this? Can I have three different values under otherwise similar conditions at steady state? Yes, it is possible that means I have got something called as multiple steady states here. So, for a given value of 5 I can realize three different possible rates why rates because rate is ultimately eta into k c a I have three different values of eta means I have three different values of rates the steady state rates possible. So, if I am doing a reaction in a CSTR or in a plug flow reactor or any reactor for that matter if I am dealing with a solid catalyzed reaction solid is porous in nature reaction order can be first order or it can be any other order. And the reaction is exothermic there are heat effects and temperature rise inside a particle is significant the thermal conductivity is small heat dissipation is relatively less under these conditions for certain values of beta what I am going to realize is that there are three different steady states possible. Now, multiple steady states is something that you already learn in reaction engineering part one for nanosothermal reactors especially the exothermic reactors. So, this phenomena is quite similar to what you have already learned in the case of exothermic or adiabatic CSTRs exothermic reactions with adiabatic CSTRs when the temperature rise gives a positive feedback and your rate also increases you have a runaway type behavior. And there is a possibility of realizing multiple steady states I hope you remember that like what you have learned in the first part dealing with exothermic reactions you have a CSTR exothermic case I will just quickly revise this non-esothermal the temperature here is different from temperature here and you have this S type curve where you have this conversion and temperature and you have some tau here and you see different states possible. You have heat generated heat removed temperature this is very quick you have to just go back and look at what you have learned before. So, the three different steady states possible heat generated curve intersects the heat removed line for temperature of the reactor this is what we have already learned. So, why I am referring to this is because this phenomena that you are seeing in inter particle diffusion it is quite similar to what you see in a normal exothermic reaction. So, here also the effect is because of the or this happens because the heat effects exothermic reaction a similar thing is happening in the case of reaction catalyzed by solids where the heat is significant temperature rises there and because of that the rate constant is increasing and the rate of reaction is increasing. And I realized multiple steady states now of course it is shown in the graph of phi 1 versus eta whereas I use different state variables for CSTR tau versus x n t, but the reason is similar. So, whatever you have learned there is applicable to non-isothermal reactions with inter particle diffusion as well. You are not going to spend much time analyzing the equations and why you are getting steady states and all because it is quite similar to what we have learned before. Another important thing is in these types of curves this is S curve that you are seeing this happens at only large values of beta not for all values see look at relatively small value of beta you do not see multiple steady states only at very large values. Now, these two are normally stable steady states and this is unstable steady state this comes again quite similar to a CSTR multiple steady state problem. Unstable means you cannot realize it because it is difficult to have open loop or you can have a steady state being stable there though mathematically you have a solution stability point of view it is not a stable solution this is something that you have learned in control theory as well there are stable solutions and unstable solutions. So, in reality if you want to operate a particular say reactor or system it normally gets stable at a stable solutions, but unstable solution you have to have a proper control to get to that particular unstable solution because system has a tendency to go away from it and probably you have learned all this in CSTR exothermics reactions. So, just remember this again you have S curve unstable solution and there are two stable solutions. So, it is quite possible that for a given value of phi you have one solution for eta which is close to 1 that means there is no effect of particle diffusion, but it is quite possible that you may get very large value of eta and a rate of reaction would go up and that depends on the dynamics of the system how you start your reactor what is the initial condition are there any fluctuations. So, that you go from one steady state to another steady state this is another practical constraint in this if you have very large value of eta the internal temperature will go up like anything these are theoretical calculations considering that high temperatures are not affecting the catalyst, but there is always some practical limit on the temperature at which the catalyst can operate actively can work actively because at high temperature you learn in deactivation that at high temperature this small metallic particles have tendency to come together and they will agglomerate and the surface area of the metal expose the reaction environment will go down and activity will go down it is called a sintering. So, every catalyst has upper temperature limit. Now, what we are showing here is not considering all those practical limits assuming that the catalyst is active at any temperature and calculating this eta, but then what happens is at high temperature catalyst starts deactivating the rate constant would go or will change. So, the equation is no longer valid or the value of rate constant that we are using is no longer value. So, all these effects are there keep that in mind. So, you may not realize that steady state at high temperature or high value of eta you may not realize you may realize realize or may not realize, but if you realize you know the meaning of it. So, this is about when the reaction takes place in solid catalyst exothermic reaction and temperature effects are significant. So, I think this discussion is good enough as far as knowing what will happen if you have exothermic reaction, but then sorry one more point if you have endothermic reaction will you realize multiple steady state? Do you see multiple steady states endothermic CSTR you do not see that why because endothermic reaction effect is exactly opposite I am talking about adiabatic case. So, there are no heat loss to the surrounding in both the cases in CSTR also and in the case of particle also. Of course, some amount of heat is lost till you see multiple steady states, but that typical effect of reaction heat is to be is should be evident. So, now for exothermic reaction is a rise in temperature rise in temperature would cause rise in rate. So, this is a positive feedback given rise in rate would cause increase extent of reaction increase heat and this is a vicious cycle till the reaction stops, but then this positive feedback gives rise to multiple steady state. This is a qualitative explanation there is so much that one one can there is already there is so much theoretical development in this area, but let us not get into details of that, but what is necessary this positive feedback will you see the positive feedback in the case of endothermic reaction. No, because as the extent of reaction increases the temperature would go down and that will cause the drop in rate of reaction and that will not give a positive feedback and this runaway type behavior. So, endothermic reaction you will get only one stable steady state endothermic reaction you will get only one stable steady state. So, in this plot for value of beta then 0 you will go in this direction and less than 0 that is when you have endothermic reaction you will go in this direction. So, there is no possibility of multiple steady states in this zone you see multiple steady states only here less than 0 means there is a drop in temperature there is not rise delta t is negative. We are talking about a magnitude here delta H is a magnitude of reaction, but then for endothermic reaction it will be negative, because delta t that is the temperature delta t max is going to be negative. So, that is how we look at it endothermic reactions will not give rise to multiple steady states will not give rise to eta greater than one only for endothermic reaction. There is one more possibility you can imagine this situation normally does not happen, but then there is a possibility that eta can be greater than one for negative order reactions. See it is like this even if it is isothermal this is isothermal. So, your concentration goes down inside a particle concentration going down is not good for many of the reactions, because the rate of the reaction goes down, but for reactions with negative order reaction see r is equal to minus k c a raise 2 minus 1 or divided 1 1 upon c a rather. In that case the drop in concentration will give rise to increase in rate. So, the rate inside is going to go up or is going to be higher than that calculated at external surface. So, there is a possibility that eta can be greater than one. Unusual cases normally you do not come across such situations, but you can imagine such scenarios where there is a possibility of eta being greater than one, but in reality for exothermic reaction people have observed specially say ammonia synthesis or methanol synthesis where you are dealing with exothermic reactions porous catalyst you see all these effects that is eta can be greater than one, but not for negative order or not for there are not many cases where we come across negative order reaction, but theoretically it is possible to get eta value greater than one if reaction is negative order with respect to reactant. So, we have looked at heat effects now there are certain things that we are not or we have just made a passing remark while we discuss various aspects. One thing is about the geometry. So, far we have looked at spherical particles, but there is a possibility that you may come across a cylindrical particle because pellets can be cylindrical as well or you may have a slab possible. You may have a monolacid this is your reactor tube and you are coating the tube with catalyst the walls are coated by the catalytic material which is porous in nature. What is porous? The catalyst is porous wall is porous catalyst is porous and then your reactants are flowing this way and on their way they will get diffuse through this and the reaction will take place. There are many possible scenarios and everywhere we need to consider the pore diffusion effects as long as we are using a porous catalyst. Now what is the difference between what we have learned here and these geometries? The only difference is the derivation and the expression for effectiveness factor and Thiele modulus. I have already told you that there are steps that we can follow 1, 2, 3, 4 to get the expression for effectiveness factor or Thiele modulus. What are these steps? You have to write component material balance use proper boundary conditions a differential equation with boundary conditions gives you the concentration profile inside a pellet let it be spherical, cylindrical, slab whatever and this concentration profile helps you get the rate at the external surface and rate inside a particle. Concentration profile gives the rate inside a particle external surface you know like r is equal to k c a s raise to n whatever. So, effectiveness factor is the rate calculated based on the flux divided by the rate at external surface. So, this numerator that is rate calculated on the flux will be obtained by the concentration profile and this concentration profile comes by writing the component balance for different geometries and we already done it for spherical particles. You can do it for say slab for example, I am not going to derive this, but I can imagine a slab with thickness say l or 2 l there is a symmetry here of course, the slab is like this, but that is insulated from all the sides only the diffusion is taking place in this direction. Then I can consider a differential element here like what I did for spherical particle and then this is my x. So, this is 0 and this is l this is x and this is x plus d x. I can write balance for this differential element I get the differential equation which will be of the form say d 2 c by d x square d e something like this minus what k rate constant in proper units into depending on the order of reaction k c a raise to n is equal to 0. Something similar and then I write boundary conditions again at x is equal to 0 you have the flux 0 or concentration finite and you know the external conditions use those boundary conditions you have a set of equations. So, solve them either analytically or numerically and you get a concentration profile you get a concentration profile. How do you calculate eta you have flux into area divided by rate at the external surface this is quite straight forward area you know this into whatever we flux comes from the concentration profile. It is this gradient d c by d x or in dimensionless form it is d psi by d lambda already I have got a concentration profile get a gradient put it here and you get a value of eta. So, that is the general procedure. So, I have just shown you what happens in the case of slab similar thing for cylinder use cylindrical coordinates here you are assuming that this is all non catalytic diffusion is taking place only like this. So, you can write this for any geometry where the pore diffusion is significant people have done it for a pore like it is a cylinder through which there is a movement particle sorry not particles molecules there is a diffusion taking place and this is very small pore, but there is a catalyst on the wall something similar to I what I showed just now where you have catalyst coated on the wall of the reactor. Now, consider this is the reactor this is small pore, but consider this is tiny reactor where you have the diffusion taking place no convection the diffusion taking place. I can write similar equations differential balance in out get a profile get a profile get a gradient put it here and you get a value of eta. So, do not get confused if you have a different geometry the remember the procedure that is more important concept is important. So, for 3 different geometries the first one of course is spherical you have the values of phi first sphere phi is equal to r root of k 1 rho c s a divided by d e for first order reaction the moment I write k 1 this first order reaction then for cylinder if you follow same procedure what you will realize is phi is equal to r by 2 now. So, if the 2 comes out of that what surface area volume and all the relationship with radius here came probably because you have 4 third pi r cube 3 came there. So, this is cylinder first order slab l now there is no r for the slab k 1 s a rho c divided by all in the square root. So, look at this factor in the square root it is same throughout it does not change and nothing to do the geometry only the part here is different from geometry to geometry. And once you have that then the relationship of phi with respect to eta will be similar for the sphere cylinder and slab. So, irrespective of any geometry the relationship would be similar that means as phi increases eta would go down. So, only it is a value of phi that would matter for the geometry. Once you get phi then you have the concentration profile in the form of phi a typical relationship and this relationship will not change from geometry to geometry. So, for every geometry you will have the same relationship between eta and phi which is 3 by phi whatever cottage phi into phi minus 1 similar of course. Now, let us look at some possible examples in the case of like we are not at solve any solve problems. There are many problems in many text books Fogler's book there are many interesting problems you can try and solve them the solutions are also given just one of those problems will have look at it again we are not going to solve it completely, I will tell you how to solve it. So, there is one problem in Fogler's book which is again for a simple reaction a going to b the first order reaction irreversible spherical particle right. Now, of course, you have platinum coated on some support particle or pellet whatever you say it can if it is small then it is called as particle and if it is specially made to suit the requirement or rather to meet the requirement of what you need in the fixed bed reactor it is pellets. What they have given you is the reactant the reactant concentration the reactant concentration C A right half way between external surface and center of the pellet is equal to one tenth of C A S that is external surface concentration right. So, you have C A by C A S is equal to 0.1. So, C A by C A S is equal to 0.1 the concentration at the external surface C A S is equal to 0.001 gram moles per d m cube the diameter of the particle is 2 into 10 raise to minus 3 C m that is centimeters is the diameter of the particle is always given the diffusion coefficient effective diffusion coefficient is 0.1 centimeter square per second see the value order of magnitude we not at looked at these values numerical values. So, just try and get a feel for these numbers this is a data given to you right and from this data what is asked is now what. So, this is your spherical particle this is center you know the concentration half way through you have been given this data and what is asked is the concentration the reactant at a distance 3 into 10 raise to minus 4 centimeters from the external surface. The data given is half way through that is if the particle diameter is 2 into 10 raise to minus 3 the radius is 1 into 10 raise to minus 3 half way through is 0.5 into 10 raise to minus 3. So, this distance is 0.5 into 10 raise to minus 3. Now, I am looking at what I want to calculate is the concentration at a distance 3 into 10 raise to minus 4 centimeters it is a very simple problem what will you do you know the equation for the concentration profile what is it C A divided by C A as psi is equal to 1 by lambda sin h psi into lambda divided by sin h what is it sin h phi sorry psi sorry this is phi. So, this equation is given to you or is known to you rather right. So, from this equation you know it is given 0.1 for lambda 0.5 sin h lambda 0.5 divided by sin h phi. So, you have this equation only phi is unknown you can use trial and error to get a value of phi which will be close to say 6 here right. So, once you know the value of phi again what is asked is what is the concentration at the at a given radius. So, again psi is equal to C A divided by C A as is equal to C A divided by what is it 0.1 sorry it is the concentration at external surfaces 0.001 sorry 0.001 is equal to 1 by lambda sin h say phi which is 6 into lambda 1 by lambda sin h say phi which is 6 into lambda divided by sin h 6. So, what is lambda 3 into 10 raise to minus 4 from the external surface. So, lambda is equal to R by R external surface that means R minus 3 into 10 raise to minus 4 divided by R what is R R is 1 into 10 raise to minus 3. So, 1 into 10 raise to minus 3 minus 3 into 10 raise to minus 4 divided by 1 into 10 raise to minus 3. So, this becomes 0.7. So, substitute 0.7 here you get a value of C A. So, after substituting the C A is about 2.36 into 10 raise to minus 4 moles per d m cube I will complete this solution in the next class.