 Today, we are looking at some practice problems concerned with non catalytic gas solid reactions. First one is on fluidized bed. Quickly draw out fluidized bed looks like. So, let us look at an example. Let us zinc sulfide plus oxygen giving you zinc oxide plus sulfur dioxide. The context is that zinc sulfide is available in Udaipur region of this country and fluid bed is the contacting that is used for making molten I mean metal zinc by first roasting of zinc sulfide. Now, why is it that they use fluidized bed? Why they would use a fluidized bed? The reason is this reaction is exothermic and generates a lot of heat and you can get that heat out by putting a coil in which you put water through and then you get steam out that is steam. So, it is one way by which you can recover heat of the reaction quite effectively. So, this is one of the major reasons why people would use a fluidized bed. Now, just put it in the context let me do small calculation as to what the numbers look like. I will do a small calculation. Let us say you have sulfur plus oxygen giving you sulfur dioxide. Let us look at a plant producing 1000 tons per day of sulfur dioxide which means roughly I am just putting some rough numbers 500 tons per day of sulfur is what is consumed roughly. We all agree? Now, amount of steam that can be produced from 500 tons of sulfur I mean if you look at the heat of combustion and so on you will find that you can produce about 2000 tons of steam. On other words for every ton of sulfur that you burn about 4 tons of steam can be produced. Therefore, 1000 tons per day sulfur dioxide plant will produce about 2000 tons of steam. Now, if you put this steam through a turbine and produce electricity about 0.25 megawatt hour per ton of steam is what you will generally get 0.25 megawatt hour. So, 0.25 megawatt hour from 2000 tons of steam. So, what is the total amount of power that you can produce? It is about 500 megawatt hour per day or if you talk about 25 hours in a day just for calculation sake this about 20 megawatt of electricity what can be produced in a 1000 tons per day sulfur dioxide plant and this is what they do also this is what is done. Now, the fluid bed becomes quite effective the fluid bed becomes quite effective because the heat transfer coefficient here is typically about 200 kilocalories per meter square per hour. It is fairly large heat transfer coefficient that is one of the reasons why fluid beds are preferred particularly if you want to recover heat from the chemical reaction or if you want to provide heat to a chemical reaction also fluid bed becomes very useful because the heat transfer coefficients are quite good. It is in that context this particular problem is has been taken. So, let us look at this problem the problem says you have a gaseous you have a fluid bed I will draw it in the form of a stirred tank easier to you have gas coming in and then solids coming in. So, you have solids I will draw it from here. So, the gas and our reaction is a gas plus b solid gives you r gas plus s solid. What it says is that this reaction is under reaction control reaction control and it also says that the time required for complete consumption is 1 r 1 r and what else does it say and you have a fluid bed in which solids are fed at 1 ton per hour and then the solids are taken out through a this one solids are fed this is I will put the solids feed here solids 1 ton per hour. The question is in the next page is what is the question find the weights find the weight of solids in the reactor if gas and solids are in mixed flow. Gas and solids are mixed flow means what if this is this is x a is the conversion here and this is the solids coming in x b and x b and x a are related by their stoichiometry which you have studied already it is a mixed flow that is the important point. How do you do this problem solids are going in solids are coming out gas is going in gas is coming out and it says he wants 90 percent 90 percent gas conversion this should be 0.9 how do we do this problem. So, do this first let us recognize that these solids these solids that are coming in it reacts by the shrinking core model something that we have learnt already and this shrinking core model under reaction control shrinking core model under reaction control that we have learnt already it says what d by d t of n b if it is a single particle I am not talking about fluid bed now it is a single particle reacting this reacts as per some in this particular case it is some minus of k s sorry k s reaction control k s time c a g and it does not say it is irreversible. So, and then stoichiometry coefficient is 1. So, that also is k s time c a g and what is the surface area of interest if it is reaction control we said it is should be 4 pi r c square is it ok. This is something that we have said before and we just stating what we have known already the left hand side we said that d by d t of 4 by 3 pi r c cube times rho b this is what and this is minus of k s c a g times 4 pi r c square this is all right. Now, this we have integrated and then expressed we have done it in class. So, we do it again. So, we have integrated this and then finally, we got a result something like this I just state the result r c by r equal to t by tau r where tau r is given as rho b times r divided by k s time c a g and stoichiometric factor in this case it is 1. So, it is 1 is it all right now let us look at the problem now problem says the time for complete consumption of the particle with concentration of gas at c a 0 is 1 r the time for complete consumption of the particle with at c a 0 if it is 1 r what happens in our fluid bed what is the concentration of gas in contact with the solid in the fluid bed. We have done the experiment the data given is that if the concentration of gas this going in at c a 0 the data given is that if it is at c a 0 the time for complete consumption is 1 r it is undergoing 90 percent reaction. So, what is coming out what is the concentration of gas coming out it be 0.1 c a 0. So, this is 0.1 c a 0. So, if the gas that is coming out is 0.1 c a 0 what is the time for complete consumption if the gas concentration is 0.1 c a 0 in contact with the solid if it is c a 0 it is 1 r if it is 0.1 c a 0 what is the time for complete consumption 10 rs correct is it ok with all of you. Therefore, tau r at 0.1 c a 0 is 10 hours do we agree with this all of us. Now, the question that is in front of us is we have these solids these solids it is reacting as per shrinking core model where the r t d of the system is given by the r t d of stirred tank. This you have said correct the average conversion from a reactor where the r t d is specified is given by 0 to infinity and we said we integrate this 0 to tau tau to infinity we have done that. So, we can integrate this in part 0 to tau of 1 minus of x b times e t of d t plus tau to infinity 1 minus of x b e t of d t. And we said this term disappears it disappears because 1 minus of x b is identically 0 in the range of tau to infinity that is why we deleted this term is this point clear to all of you why we deleted this term. The reason is that when tau time is greater than tau the particle which spends time greater than tau reacts completely and that is why x b is 1 therefore, 1 minus of x b is 0 therefore, this term disappears. So, you only have to integrate 1 minus of x b e t d t given what is the value of 1 minus of x b let me write it down let me write it down here from here we get r c by r is 1 minus of t by tau r this is 1 minus of x b is it right r c by r is what 1 minus of x b to the power of 1 by 3 is it ok or 1 minus of x b equal to 1 minus of t by tau r to the power of 3. So, we the average that we are looking for is 0 to tau 1 minus t by tau r to the power of 3 and e function for a stirred tank is ok this is all right. So, this is the answer we are looking for is it all right now can you integrate this can this be integrated can the right hand side be integrated or not. So, let me just expand this and see whether it looks ok or it looks very complicated let me expand this 1 minus t by tau r minus 3 plus 3 t by tau r minus whole cube all right. So, is 1 by t bar e to the power of minus t by t bar of d t this is the integration that you have to do integral sorry integral 0 to tau tau r actually is it this can be done by all of us is there is a difficult complicated take too long now it might take a little while. So, to save some time what I have done we you can do this at home it simplifies like this I will just put this final form we should do it ideally, but you know it is not that it is too complicated. So, we can do this at home we just get this. So, where alpha is tau r by t bar alpha is tau r by t bar now what is given problem specifies that you need 90 percent of what 90 percent of the solid is to be converted. So, what is x b 0.9. So, x b is given as 0.9 equal to thrice alpha minus of 6 alpha square plus 6 alpha cube minus of 6 e to the power of minus of alpha by alpha cube can we solve this I want you to solve it here a way to solve this is take some values take some values 3 or 4 calculations all that is required to get a feel for the kind of answers clear for numbers we have to take. Let us do one small calculation all of us let us say alpha equal to 10 quickly say alpha equal to 10. So, what is the right hand side alpha equal to 10 I am just writing it down you tell me whether I have done it correctly alpha equal to 10 yes or no. So, what is the value right hand side 0.24 0.24 do we all get this 0.24. So, therefore, alpha equal to 10 is not the answer let us try alpha equal to 5. So, 3 by 5 minus what is it 3 by 5 minus 6 by 25 plus 6 by 125 minus 6 e to the power of minus 5 divided by 125 is it all right. So, what is it now. So, right hand side is 0. We all get this it is ok with everyone. So, even this is no good alpha equal to 5 also not a good answer all right let us try what alpha equal to 1 alpha is 1. So, 3 by alpha. So, r h s 3 by alpha minus 6 by 3 by 1. So, 6 by 1 plus 6 by 1 minus 6 e to the power of minus of 1 by 1. So, what is it become 0. So, r h s equal to 0.79 is it I have made some mistake is it we are closer. So, shall we now try 0.5 0.5 is 0.89 we almost got it then good. So, it is 3 by alpha sorry alpha is 0.5 and then 6 by 0.25 plus 6 divided by 0.125 minus of 6 e to the power of minus 0.5 divided by 0.125. So, that according to my friend is 0.89 it is about this is r h s. So, l h s is 0.9. So, our answer is probably around 0.5 say it again it is 0.4 is the answer is it. So, alpha equal to 0.4 my friend says x b becomes 0.9 I see you can directly solve this is it very good very good. So, let us go with alpha equal to 0.4 because it is a neat number. So, now I ask you what is the hold up of solids in the fluid bed alpha is 0.4 alpha is 0.4 what is the hold up the question is the question says find the weight of solids in the reactor weight of solids means what hold up of solids. So, w divided by f s equal to t bar s n o and alpha equal to tau divided by t bar. Now, do we know alpha do we know alpha it is 0.4 do we know tau what is it what is tau 10 hours. Why is it 10 hours because the gas is in contact with a gas of concentration which is only 10 percent of the data that is given to us correct that is why tau is not 1 hour, but it is 10 hours. So, tau is 10 divided by t bar. So, t bar is how much 10 divided by 0.4 that is equal to 25 hours. Now, what is f s it says treating at the rate of 1 ton per hour correct. So, f s is 1 ton per hour therefore, w divided by 1 equal to 25 therefore, w equal to 25 tons. So, hold up of solids is 25 tons is this clear. What we are trying to get across to you here is in gas solid reaction single pellet analysis we do experiments with a given concentration of gas, but in the equipment the actual concentration may be very different from in which you have done your experiments. You must correct the time required for complete consumption by appropriately accounting for the gas concentration which is responsible for the reaction that is what this problem is all about. We go to the next question. So, here we have a rotary kill in the handout I gave you that minus tau this is to read as l n of beta minus of x a divided by beta equal to minus tau g times alpha this minus is not very clear. So, please make that correction in your now here we have a rotary kill. Now, when do we use a rotary kill you have solids gas this is gas this is solids and then products are coming out. The rotary kill is a very popular device particularly in the lime calcination lime calcination lime calcination lime calcination around the world uses probably 15 percent of the total carbon dioxide that goes into the atmosphere comes from lime calcination about something like 6 billion tons of carbon dioxide measured as carbon or 25 billion tons of carbon dioxide is what we throw into the atmosphere these days out of which 15 percent comes from lime kill it is not a small quantity. Now, why rotary kill for lime calcination why rotary kill for lime calcination why rotary kill for soda ash manufacture if you go to factories making soda ash you will see a rotary kill. How do we explain this we talked yeah temperature required are very high and therefore, how do you supply that heat in a rotary kill that is more difficult see in rotary kills the heat that is required is actually supplied along with the raw material. For example, if you go to lime kill they will actually add lime and then coal also coal and lime go simultaneously together. So, that as the coal burns that the heat of reaction is used for the decomposition of lime. So, that means what you have to take care is that whatever you added to the raw material that does not affect or it benefits the final product. For example, coal combustion the ash actually is a part of the final product itself. So, it does not affect I did not very seriously the quality of the final product. So, in rotary kill you will generally put your fuel also along with the raw material it is not very easy to provide heat from the walls of the rotary kill because it is rotating it is not very convenient. So, what is done is with nicely insulated therefore, you do not lose too much of heat, but all the heat of combustion is actually added along with the raw material. So, you have let us say C A C O 3 plus heat giving you C A O plus C O 2, but in this case it is not the problem there is A gas plus B B solid B B solid B solid equal to C gas plus D solid. So, it is not quite similar to lime calcination can we think of any other example of process industry where gas and solid react to give you gas and solid as an example zinc sulphide to zinc oxide. Now, zinc sulphide to zinc oxide people do not use rotary kill why zinc sulphide to zinc oxide rotary kills are not common. If you go to Hindustan zinc in Udaipur it is a fluid bed quality of product in zinc sulphide fortunately zinc sulphide combustion or roasting gives you a lot of heat. So, there is no need to provide external heat source. So, that is why we indeed what we want to do is recover lot of heat from zinc sulphide roasting that is why rotary kills are not suitable for heat recovery not suitable for heat recovery. Rotary kills are very good for very large throughputs very large throughputs are what soda ash for example large throughputs. So, it is very useful. So, what is the closest example of A gas plus B solid giving you C gas plus D solid what is the closest we can think of in process industry closest cement what do they do in cement industry the cement clinker is made in this what happens there you have lime you have coal and all the fluxes and actually the clinker forms inside the kiln what is the product gases is carbon dioxide what is the feed gas air. So, air burns gives you the heat that is required for the cement production to take place. So, very good example of cement manufacture is excellent example of this reaction. Now, it says the reaction it says that it is under shrinking core with external diffusion control. Now, how do we justify in cement manufacture for example, how do we justify external diffusion control do you think it is a justifiable assumption if it is cement manufacture you will find some of you have seen the cement industry what comes out is a clinker the clinker is typically about 7 8 10 centimeters you know it is not small not 10 maybe 5 6 centimeters. So, it is get a clinker which is then powdered and then sold to us as cement. So, what you make is clinker. So, clinker has a reasonably large size now here it says external mass transfer control. So, would you think that cement manufacture would be an example of external mass transfer control what is your perception I hope you understand after try and understand if you are working in Larson and Tubro you building up a cement plant in one of these places in one of the largest you know LNT builds many of these plants. So, you might be involved in design of some of these things. So, is this question is this assumption correct or it is not a good assumption. Now, the answer is like this you will find that reactions in which there is a solid product formation external diffusion internal diffusion and reaction all are important we have talked about it when the temperatures are very high reaction kinetics is not all that important reaction kinetics is not very important it is external diffusion or diffusion through the product layer. See these are the two important layers. Now, what seems to happen in this in the cement industry is that what you add is actually powdered rock of limestone powdered coal. So, as it is going through it is actually because of the temperature it forms a clinker. So, for a substantial part it is only heat transfer to the particles which is very important only in the latest stages as because of glassy things when it forms a solid it becomes. So, for substantial part of this kiln it is actually controlled by external mass transfer that is the example that we have taken. There are better examples we can take. So, this is not a bad example although it may not be the best example. Now, we have done all these mathematics, but we want to do it again just to put it in the context what is the context you have a gas. Let me just quickly write down just put it in the context our a f a 0 times 1 minus of x a and we have b is f b 0 minus of f a 0 x a then you have c which is f c 0 plus f a 0 times x a. Then you have d which is f d 0 plus f a 0 x a and then we said that our equilibrium constant is k p it is given by what gas is p c divided by p a. I put a star to indicate that it is equilibria and we expressed all these things p c and all that in terms of concentrations we would not do it again. So, k p p c star divided by p a star. So, from our stoichiometry we get something like this correct we have done this. So, I would not go this again. So, that our equilibria is given by k p minus of theta c divided by 1 plus this we have done. So, we know what is the equilibria and how it is determined by the compositions of our choice is it what we have done. So, this representation comes from stoichiometry we have done this before we have done this here and it directly follows from stoichiometry we have done this in class. Now, rate at which the reaction takes place the rate we say this is r a dash divided by where r a dash is reaction rate per unit surface area and a s is the surface area per unit volume. This is what we have said and if it is if our reaction is what what is the external diffusion control. So, k g times c a minus of c a star times our surface area per unit volume is it. So, now we can convert this you have d f a 0 d x a by d v minus sign I have forgotten a minus I am sorry minus sign. So, it is minus minus of k g c a is 1 minus of x a c a star it is 1 minus of x a star. So, times a s times a s. So, this simplifies as f a 0 this becomes d sorry simplifies as d x a by d of gas residence time k g times x a star minus of x a cancels off and this is our a s is it all right where this x a star is what we have got here or x a star is this is the context of we have said this earlier as well the context here is that this value of theta c theta c is the products in the feed and this has a bad effect on the rate of chemical reaction because it affects x a star. And in many cases we may not have a great choice because we have to accept the fact that you know this amount we have to accept for example, calcium carbonate decomposition we said we may have to accept this theta c to be about a significant value of the total resistance all right. What is a s surface area per unit volume what we say if it is external diffusion times number of particles divided by volume. And then we said our experimentally measurable quantity is this that means the hold up of solids in the equipment hold up is actually a data that is readily available for all equipments given the hydrodynamics people will be will tell us what is the hold up. So, the hold up of solids epsilon r is readily available therefore, we are able to determine a s as s equal to 3 epsilon r divided by r something like this. So, that our equation which describes our process is this rise epsilon r by r k g within brackets x a star minus of x a. Now, if we look at this equation we find that it is this term which will determine the size of the equipment. If we can do something about this term the size can be reduced if this term is poor the size is very large. So, let us integrate this when we integrate this what do we get l n of actually we have denoted x a star this we have denoted as beta. Therefore, when we integrate this it will come out to be what will it be l n of beta minus of x a divided by beta. So, l n of beta minus of x a divided by beta equal to minus of tau g if I call this as alpha. If this is alpha it becomes something like this is it ok. Now, for the data given for the data given can be quickly calculate what is the value all of you please calculate what is alpha. Alpha strice epsilon r k g divided by r this is alpha and then beta is there it is k p minus of theta c divided by 1 plus k p. Please calculate and tell me this numbers please what is alpha and what is beta alpha is how much epsilon where is epsilon r somewhere where is that this one this is symbol is wrongly written please make a change this is a hold up is 0.15 this is epsilon actually not tau this please make that correction this epsilon this is epsilon. So, it is 0.15. So, 3 times 0.15 and k g how much is k g where are we 0.01 and what is r 0.05. So, how much is this equal to 0.09 units what are the units alpha is in what units units per second. Now, what is beta what is k p where is k p where are we k p is 5 and theta c theta c is 1 this is theta b wrongly written please make the correction. So, theta c is 1 and then 5 plus 1 is 6. So, 4 by 6 equal to 0.67. So, you what is the problem specifies that we require what is the extent of reaction sometimes mention somewhere not mentioned what is the conversion required x a desired is 0.8 x e 0.8 of x e what is x e x e is what 0.67. Therefore, x a is 0.67 times 0.8 how much is that 0.536 is it all right. So, x a is known. So, you can find out what is gas residence time tell me l n of beta 0.67 minus of 0.536 divided by 0.67 equal to minus tau g times alpha is how much. So, tau g equal to 17.8 seconds. So, what this is saying is that the reactions that is controlled by gas film does not require much residence time. So, this is the point that we all should appreciate. So, it is something else which requires lot more residence time. Therefore, if you design on the basis of the gas phase residence time you will probably get a kill which is quite small. But that would not do your job because the kill has to do lot more job because the clinker has to form the cement has been formed, but it is be converted to clinker now that may be a much slower process. That is why the kill is much longer that is the point that is trying to get across to you all right. Let us go forward what is solid residence time solid residence time tau s is what hold up of solids which is v times epsilon r divided by the flow of solids what is solid is b correct f b the solid the volumetric flow of solids may not change f b 0 divided by rho b may not change very much. In the sense that the solid density at the feed and solid density the exit may not be very different actually in this case data is not given. So, we take the volumetric flow of solids as simply equal to volumetric flow of solids at the feed which is f b 0 divided by solid density which is 50. So, what is the volumetric flow? So, what is the volumetric flow of solids density is 50 and what is f b 0 3.6 multiplied by 0.7 divided by 15 is it all right with everybody all right. So, epsilon r is 0.15 and then divided by 3.6 multiplied by 0.7 divided by 15. So, we still what is the volume of the equipment 5 0 I am sorry 5 0 ok. Now, what is the volume that we still do not know how do you find volume gas residence time multiplied by the gas flow. So, v naught times tau g is. So, what is v naught? So, f a 0 divided by c a 0 f a 0 is 3.6 c a 0 is 0.02. So, 3.6 divided by 0.02 may not change very much multiplied by tau g tau g is what 17 seconds you told me 17.8. So, how much is volume? What is the volume? How many cubic meters? So, you have divided by 3600. So, how much does it now? It looks much better now what is the answer 0.89 cubic meters all right 0.89. So, what is the solids 2.6 2.67 watt solids is 2.67 seconds. So, what we are trying to put across here is that the gas control phenomena is does not occupy much volume that is the point we are trying to get across here. So, you told that the volume of the tank depends on k g. It does not see because that reaction is not occupying too much of volume. All the rest is for converting the cement into clinker that is what takes most of the volumes. Is it all right? Shall we go forward? Yeah. That is why I am saying what we have looked at here is only the part that is controlled by gas external film. So, there is the fact that we provide a much more volume is because that form a clinker has been converted to clinker. It has to you know that is what takes a lot of space that is what I am trying to say. That is not in this problem we will put another problem to take care of that. If this one we are only looking at a small part of the whole exercise showing that this gas film control is not such an important thing as far as this particular problem is concerned. So, that is the number that comes out of this particular problem. We will do that also a little later. Third this exercise is C plus H 2 O giving you C O plus H 2. This is Q 3. So, all of you have seen this. I am sure you know there is so much talk about gasification of biomass. And what happens in gasification of biomass is that you expose the carbony emissions materials to a high temperature with steam. When you expose it to a high temperature with steam you get carbon monoxide and hydrogen. Both carbon monoxide and hydrogen are very valuable materials. In fact, there is still lot of work goes on around the world to somehow perfect this reaction. This reaction is endothermic. This is an endothermic reaction and the thermal about 25 kilo calories per mole is the kind of energy that is associated with this reaction. It is not small. The heat effects are not small, but the products are very valuable. The products are extremely valuable both carbon monoxide and hydrogen are very very valuable products. And therefore, this is of great interest all over the world people are trying to look for various kinds of catalysts to incanses rates of reactions and so on. Nothing much has really happened from biomass from biomass. If you look at coal, coal is not just carbon it also has hydrogen. So, if you look at coal I mean the situations are a slightly better situations are much better for a number of reasons. What happens is that 25 kilo calories per mole is the energy that is required for this reaction and you have to get this energy from somewhere. Then only you can drive this reaction. You ask yourself how will you provide this energy? How do you provide this energy? Yesterday we looked at what is called as a reactor and regenerated system. In fact, the context of that and we said solids are moving between the two. We mentioned that yesterday we did a problem also to illustrate. Now, looking at this and looking at this problem of coal, coal gasification can you think of a way by which we can do this coal gasification little better. Coal gasification are really not taken off all over the world. It is 50 years later we are still struggling with a good coal gasification technology. What do we do? Do we see some solution by there is in the two reactor system? In the petroleum industry all of us have seen there is this cat cracker. What happens in a cat cracker? Cat cracker there is this reactor and there is this regenerator. From the reactor the solids descend by gravity. From the regenerator it goes up by pneumatic conveying. What happens here? This is the reactor, this is the regenerator. Here catalyst gets deactivated. Here catalyst is regenerated. So, this is the regenerator, this is the reactor. Now, you want to conceptualize the same thing for the case of gasification of coal. What is the object here? The object here is gasification of coal. In this reactor gasification should occur, which means C plus H 2 O should go to C O plus H 2 O. For this you need heat. So, what you would like to do? You want to burn coal here. You will burn coal and because of that heat the solids here you want to take it there. So, that the heat the solids serve as the heat carrier for the gasification reaction. You understand? Now, this technology itself has not really taken off. It is not come to a stage where we can think of coal gasifications. But if you can do this, if you can separate the generation of heat and supply the heat by this process, then the advantage here this gas the calorific value is very high. Because you do not allow the combustion, this is air. So, this is lot of nitrogen here. This nitrogen does not get in. Otherwise, lot of nitrogen gets into the combustion gases and your thermal value is very low. See, most of this what is called as biomass gasifier that is running around the world. The thermal value will be around 800 to 900 kilo calories per cubic meter. 800 to not sometimes even less. It is a very lean gas. And therefore, it is not very suitable for various types of engines unless you have a substantial amount of additional fuel like diesel. You have to put some diesel then only it will work as it will not work. By itself, you are not able to use these gases in combustion IC engines because it is just too lean. See, this is where the technology is stuck for the last number of years. We do not have a way by which you can produce a gas which is got sufficient thermal value. And that is why coal gasification programs have more or less sort of not going forward. Now, recently thanks to a huge increase in the cost of crude oil, the interest has come back again. In India for example, I mean when I joined this department some 30 years ago, it was not very important. Now, it really depends upon the cost of fuel oil. I mean what is called crude oil. Now, the crude oil cost is so high, the tremendous amount of interest around the world. But still, we do not seem to have a good way by which we can carry solids and heat to the reactor. There is a problem we are facing. Now, the question here what is this question? This non catalytic reaction is carried out in an isothermal enclosure with large quantity of steam at temperature T. Reaction is endothermic and reversible. Set up a model, express conversion time relationship for the shrinking particle. Now, we have a shrinking particle. What do we have? We have a particle. We start with this particle and now this particle becomes this as the reaction proceeds. How do we handle this problem? We write this D by D T of N B equal to what? Some R B times S. There is no change. What is this S? S would be the reacting surface. What is that reacting surface? When you start, this radius is R. As the reaction produce, it becomes R C. So, in other words, the reacting surface is always exposed to the gas. In the shrinking particle, the reacting surface is always exposed. This is C A G and we are assuming that there is no external gas film. I think this is under reaction control. Is that right? It is not mentioned. So, at least the ash is not there. There is no product layer. So, either it is controlled by chemical reaction or it is controlled by external mass transfer or both. We do not know. Since nothing is specified, we should consider both and then bring it together. We have done that also. So, let us first look at reaction control. So, what is our N B? It is 4 by 3 pi R C cube times rho B times D by D T. The right hand side is what? If it is reaction control, K S times C A G, our stoichiometric factor if it is there, it is probably 1. I will put a negative sign to indicate that it is getting consumed multiplied by 4 pi R C square taken to account the surface area. Is this alright? Now, let us recognize that this situation is very different from the situation we considered earlier. The earlier situation was there was an unreacted core here. The gases were diffusing and we said when this reaction control, there was no resistance in the product layer. That was what was the situation we had considered earlier. Here, that is not the situation. The situation here is that the particle itself shrinks. Therefore, at every instant of time, the unreacted surface of the particle is exposed to the gas. That is the difference between the two situations. So, can we take this forward now? So, the left hand side is 4 pi R C square rho B D R C by D T on the left hand side. The right hand side is minus B times K S times C A G times 4 pi R C square. So, 4 pi R C square disappears. Yes or no? So, this we can integrate and we will get something similar to what we have got before. So, R C by R equal to T by tau. I put an S here to indicate the shrinking particle where tau R S is rho B times R divided by B times K S times C A G. Is it alright? What we have done? Now, we have been talking about shrinking core models and then also talked just now about shrinking particles. Now, we would like to look at this whole issue of shrinking particle in the instance of external diffusion control. Now, why we are looking at external diffusion control is that this could be one of the controlling mechanisms that you might encounter. And therefore, we should have appropriate formulations in our hand. So, what I have got here is a particle of initial radius R which is undergoing sort of a burning let us say and initial radius R. And as the reaction proceeds the particle is shrinking because combustion of coal for example, is a good example. So, we want to see how we can relate the extent to which the combustion is taking place to the fundamental parameters of the process. So, the rate at which the particle is reacting is the rate at which if it is an external diffusion control we said K G times C A G is the rate of supply of material. And then 4 pi R C square is the area over which the supply takes place and B is the stoichiometric. So, we have a A sorry A gas plus B B solid gives you products this is your reaction. So, B is the stoichiometric. So, this is the representation at of how the particle is reacting and how it can be related to the mass transfer coefficient the composition in the external composition. And this is the 4 pi R C square is the surface area over which the reaction takes place correct. On other words the area over which the diffusion is taking place is 4 pi R C square and as the particle is shrinking this R C will keep decreasing and to that extent this quantity will keep on decreasing. So, we have to take into account the effect of this decreasing surface area on the reaction rate that is what we want to do right now. Now what is N B we know that N B is 4 by 3 pi R C cubed rho B. We can differentiate the left hand side and notice that many differentiate the left hand side R C square gets cancelled. So, that you get rho B R C by D T is minus B times K G times C A G. Now notice here that this C A G is assumed to be this C A G concentration assumed not to change as the reaction proceeds. So, that you do not have to worry about changes in C A G of course, we will consider a situation where we take this changes effects into account shortly, but for the moment we are formulating for a situation where the composition of gas external to the solid is not changing. Now there is some point also we must bear in mind this particle is it is burning let us say therefore, it is decreasing in size. In many text books how it is described that because the core is shrinking that is correct therefore, to call it R C is some people do not put a subscript R C they just call it R showing that you know the particle shrinking yes and therefore, at any time any instant of time the radius of the particle R and not call it as R C. The term R C is used for the case of a particle which is of unchanging size when the particle of changing size people prefer to use R instead of a subscript R C. I have written the same subscript R C to so that you know we can we can continue with the same nomenclature between unshrinking particle and shrinking particle. So, what we have for a shrinking particle under external diffusion control we have rho B times D R C by D T on the left hand side minus of B times K G times C A G. So, this describes how the size of the particle is changing with time B is a psychometric coefficient K G is mass transfer coefficient C A G is the composition of the gas external to the solid and assume to not change as the reaction proceeds because it is in very large excess. Now, what we know from our basic mass transfer studies is that Sherwood number Sherwood number which is defined as K G R C by D where K G is the mass transfer coefficient R C is the particle size at any instant of time and D is the diffusion coefficient of gas from external to the solid surface. Now, this Sherwood number is known to be equal to 1 plus a constant times Reynolds number to the power of number n typically 0.8 n equal to 0. Typically this n is equal to 0.8 or so. Now, this is something that we know from our mass transfer literature showing the Sherwood number which is K G R C by D is 1 plus a constant times R C to the power of n n is typically 0.8. Now, if you are looking at flow over a solid where the Reynolds numbers are very low then we can say that Sherwood number is K G R C by D equal to 1. What are we saying? What we are saying is that for a shrinking particle the Sherwood number equal to 1 may not be a bad assumption if the flow around the particle is laminar flow. So, suppose we look at a situation of very of lower Reynolds number. So, the Sherwood number K G R C by D equal to 1 is applicable then what we can do is that in this equation in this equation rho B D R C by D T equal to minus a B K G C A G. This K G can be replaced by from this equation K G R C by D equal to 1 or K G equal to D by R C. So, what are we saying now? What we are saying now is that this equation here this equation here which tells us what is the rate at which this radius of the particle shrinking particle is changing it is determined K G appears on the right hand side this K G is actually equal to D by R C this is what we are saying K G equal to D by R C. Therefore, we are able to replace K G as D by R C here on the right hand side. So, that we get rho B D R C by D T is minus of B D by R C times C A G. Now, notice here notice here is that the rate of change of particle size D R C by D T now depends upon D by R C this dependence this dependence is what is interesting when particle is shrinking and it is under external diffusion control. We just now little while ago we derived and shown that for a shrinking particle under reaction control the form of the of the final result is the same whether it is shrinking particle or whether it is an unchanging particle size in both cases the form of the expressions are the same. But in the case of external diffusion control because of this Sherwood number equal to 1 result we have this dependence on R C as can be seen from this equation. Now, we can integrate this I have done the integration then finally, you get this result that rho B this is the result we get this result shows that this is the result which we can simplify further which I have done here. What does it say what it says is that 1 minus of R C square by R square is T by tau F S where F refers to film diffusion S refer to shrinking particle. And what is F S that means the time for complete consumption the time for complete consumption of the particle is rho B R square divided by twice B times D times C H E. Now, this result is slightly different from the result we got when we talked about unchanging particle size. So, we will we want to look at that result carefully, but before we do that let us also recognize that R C by R is 1 minus of X B therefore, the same result the same result can be written in this form showing that if there is a shrinking particle. And we want to understand the extent of reaction X B then we can write our result in this form where X B is the extent of reaction T is the time of reaction tau F S is the time required for complete consumption of the particle. So, for the case of external diffusion control shrinking particle we find that the time required for complete consumption of the particle is given by this result. This is an interesting result that we must bear in mind because it is slightly different from the result that we had seen earlier just in the in since the context is important and just drawing your attention to what we had done a little earlier. What did we do a little earlier? We talked about particles of unchanging size where we showed reaction control film diffusion control and ash diffusion control. And then we derived the time required for complete consumption of the particle for reaction control as rho B R by B K S C A G. And then film control we said is rho B R 3 B K G C A G and then for ash diffusion control rho B R square 6 B D C A G. Now moment we went from unchanging particle size this is for unchanging particle size unchanging particle. Let me write here unchanging particle size particle size. Now if it is a shrinking particle we have shown that also that it is a reaction control this is the result for film diffusion control this is the result. Now frequently our interest is to be able to tell by looking at the results what is the likely way of discerning the controlling mechanism. We have talked about it I mean as a part of the chemical engineering program. But we will just draw attention to what we all know and put it in the context of trying to understand how these reactions take place. Now suppose for example a reaction is taking place and we know that if it is a reaction control then if you change the temperature if you change the temperature you will find temperature has tremendous impact on the rate of chemical reaction. You will find the effect of temperature would be very strong if it is a reaction control. Now if it is a film diffusion control we said we know that K G the mass transfer coefficient is a very strong function of the velocities or the hydrodynamics of the conditions under which the reaction is taking place. Therefore if you do experiments at different flow velocities we can discern whether the changes are due to external diffusion or due to reaction. Because by changing temperature you can determine whether reaction is the controlling regime by changing velocities we can determine whether mass transfer external mass transfer is controlling. And since tau d this ash diffusion depends upon square of the particle size we had said at an earlier time that by changing the particle size say particle twice thrice and so on we can determine which whether ash diffusion is a controlling regime. Now we know that if it is a shrinking particle then there also the dependence of the time for complete consumption depends on square of the particle size. So there are two kinds of situations one ash diffusion also depends on square of particle size the time for complete consumption similarly the time for consumption of a shrinking particle under external diffusion also depends on square of the particle size. Now the question is how do you distinguish between the two mechanisms say ash diffusion control unchanging particle size film diffusion control shrinking particle the answer is very simple. Now you can by looking at the situation itself tell whether it is a shrinking particle or whether it is an unchanging particle. As an example let us say you are burning biomass. Now burning of biomass is a shrinking particle the reason is biomass has very little ash 1 percent 2 percent and so on therefore biomass is a good instance of a shrinking particle because the ash is very low. Therefore simply by looking at the situation itself you will be able to tell that it is a shrinking particle and therefore the question of ash diffusion control does not even arise because there is no product layer at all. So what I am trying to put across to you is that by looking at the physical situation by looking at the data that is in front of you carefully you will be able to discern the controlling mechanisms. Of course the way to do it is that you should do experiments at different temperatures you should do experiments at different flow velocities and you should do experiments at different particle size. Therefore you have to do a number of measurements and looking at those measurements you will be able to discern where you are what is the extent of control that you are seeing from different controlling mechanisms. Having said this having said this we want to draw our attention to some an important issue that we will face in days ahead and that is that is energy that you and I require. Now as you all know as you all know that fossil fuels are creating various kinds of difficulties to our global environment. The reason being of carbon dioxide and so on. So anything that we can do with biomass would be a great value. But biomass is limited by the extent to which we can grow biomass because of limitations in agriculture plantations and so on. Therefore maybe there is a time for maybe for a number of years from now maybe 20, 30, 40, 50 years we may have to depend on coal. Both gasification of coal and gasification of biomass it is a it is an area of great interest around the world because this provides energy for our daily needs. Now whether it is coal or it is biomass the gasification technology what is that gasification technology. The gasification technology is that you react carbon with steam to give you carbon monoxide and hydrogen. This reaction C plus H 2 O equal to C O plus H 2 is a well studied reaction. It is endothermic its heat of reaction is 29.9 kilo calories or 1000 Kelvin. Its free energy change at standard conditions is minus 1.9 or K P is about 2.6. In other words it is a reaction where the equilibrium constants are low it is endothermic and therefore you must choose reaction conditions which are appropriate for these kinds of thermodynamic parameters. But there is a more important feature that you must recognize. What seems to happen is that if you try to gasify either coal or biomass using a mixture of steam and air or in other words you do this reaction let me just put it down here to this reaction something like this. So, you have a enclosure let us say you have an enclosure and then you have say biomass of coal is going in by and then you put air and steam at an appropriate temperature. So, that here you get gases the reaction is C plus H 2 O giving you C O plus H 2 and then plus because it has air this reaction will also take place it gets C O 2 this also will happen. Now, what seems to have been the major difficulty around the world about gasification technology is that moment we use air to generate the heat this is exothermic correct this is endothermic. So, that means you are generating heat by this reaction. So, that it can be supplied to this reaction that means internally you are using the exothermic heat of this reaction to supply the endothermic heat of reaction for reaction one if I call this reaction one call this reaction two. But in that process of course there is a great synergy here understandably, but the fact is that since this oxygen is coming from air this gases contain lot of nitrogen. The gases that come out contains lot of nitrogen and the experience around the world is that something like 75 percent is nitrogen. Therefore, this is a very lean gas it is very lean gas the thermal values are very low may be 6 to 800 kilocalories per cubic meter. And it is so low that make the thermal thermodynamic efficiencies we can reach in our engines etcetera in our boilers etcetera are also not very large because it is very lean gas often we have to supplement some other gas and so on some other fuel and so on. Now, therefore, the problem that is of interest to us of course it is a great interest all of us that if you can somehow separate the exothermic heat generation this is exothermic this is endothermic. So, if you can separate the exothermic reaction and then move the solids to supply heat for the endothermic reaction that means by separating the endothermic and the exothermic reactions and making the solids serve as heat carrier we are able to achieve a very I mean high value high value high calorific value calorific value gas. So, we are able to achieve a very high calorific value gas by doing this kind of heat supply using the heat carrier that means you have two fluid as beds and then the fluid as bed here this serves as the heat carrier into the endothermic reaction and this that material comes back by gravity. So, you have a pneumatic transfer and gravity transfer of solids. So, this solid transfer ensures that the heat is supplied and then therefore, you are able to get a high value of high thermal value for the gas. Having said this this whole idea has worked very well in the cat cracking industry it continues to work very well in the cat cracking industry, but in biomass gasifications and coal gasifications these ideas have not been very successful yet a great deal of work needs to be carried out before we are in a place to produce high calorific value gas from coal and biomass. Having said this our present interest in this exercise is to be able to set up the stoichiometry set up the equations to understand how the equilibria affects the rate of chemical reaction because after all ultimately our rate functions are required to be able to tell how these reactions will proceed in a reaction equipment. Now, the example we are trying to talk about is that there is a continuous input of this material it can be coal it can be biomass of course, coal means it has so many other things as well I am just simplified representation. So, we are talking about a continuous process in which a material is coming in it is reacted with steam at a very high temperature typically 1000 k. So, that you have carbon monoxide and hydrogen. So, it is this problem we want to represent appropriately. So, that we can understand from the stoichiometry you know how the equilibrium constant affects the reactions and so on. So, I have written the stoichiometry here. So, you have a b c d or the gases coming a is moisture and sorry steam and b is carbon we are taking this carbon what we mean is coal or biomass and now c and d are gases c and d are gases. So, I have written our stoichiometry in this form showing that because the heat of reaction I mean is endothermic it is 29.9 kilo calories and it is very the equilibrium constants are quite low it is only 2.6. Therefore, we will have to take into effect the equilibria on the reaction rates. We will do that shortly let us write the stoichiometry now. So, now that we know this is the stoichiometric table we can return what is concentrations all this concentration C A C B C C C d can now be written in terms of the extent of reaction which is x. Now, when trying to do that what we want to understand is that we have a gas law as per gas law we have written the gas law we have done this before let me do it again for you gas law is what gas law is v by v 0 equal to p 0 by p t by t 0 z by z 0. And f t by f t 0 we know this f t by f t 0 p 0 by p t this is gas law therefore, if you want to find concentration you simply have to say what is the molar flow of component A divided by the volumetric flow of component A. And what is the volumetric flow because f t by f t 0 you can see from here when you add up all this you can say total comes out to be if this total is twice f A 0. So, I am just adding it up here let me just add it up here. So, that we all understand let me just write here. So, the total gas coming in is what f A 0 is coming in and then what is going out going out is f A 0 times 1 plus x A. So, you can understand all the inputs are f A 0 nothing else is coming in output has f A 0 1 plus x A that means, there is an increase in volume this is in this is out this is f t this is f t in and this is f t out I call this f t 0. So, f t by f t 0 is 1 plus x A this is all I am trying to say. So, since f t by f t 0 is 1 plus x A v by v 0 v equal to v 0 times 1 plus x A that is what I have written here. So, what we have done is that we have been able to express concentrations of the gases A B and C in terms of the compositions at the inlet and conversions which is measurable all these conversions are measurable. So, we have C A C B C C C D not C A C C and C D in terms of inlet compositions and conversions having done that we can now recognize that the equilibrium constant for this reaction what is the equilibrium constant for this reaction k p k p is p c p d divided by p A we know this correct. And what is the rate at which chemical reaction takes place we also know the rate at which chemical reaction takes place under external diffusion control it is k g times the concentration driving force what is the concentration driving force C A minus of C A star what is C A and what is C A star this is the particle which is swinging C A G is outside and then C A is on the solid surface at any instant of time what you finding is that this is the driving force this is C A star on the surface. And then this C A we said it is not going to change therefore, we can call it a C A G or C A star C A 0 I mean. So, your driving force is C A G minus C A star C A G is not changing and therefore, in terms of our conversions I have written it as C A 0 x A star minus of x A correct what is C A we have written C A as. So, we will have to divide this by 1 plus x A 1 plus x A. So, is that clear what are we saying what are we saying is that our driving force just let me just I will just come back to in a minute what is our driving force C A G minus of C A star correct. So, that is what is we have just now written here C A minus of C A star is given by this C A 0 1 minus of x A 1 plus x A therefore, this C A this C A star and C A it is known to us because we have written all these things in terms of x A. So, what we are trying to say here is that C A G minus of C A star can be written as C A 0 x A minus of x A star this is something we know because C A are all known. So, divided by let me put this 1 plus C A equal to C A 0 times 1 minus of x A divided by 1 1 plus x A therefore, C A star equal to C A 0 times 1 minus of x A divided by 1 plus x A I will call this as C A star and what is C A what is C A C A equal to which is already mentioned the C A 0 times 1 minus of x A divided by 1 plus x A. So, what we are saying now is that what we are saying now is that R B R B which is equal to minus of K G times C A G minus of C A star. Now, this can be written as in this form therefore, it is equal to minus of K G if this C A G is changing then it will be written as C A minus of C A star where C A is given by where C A is given by this expression C A star is given by this expression. If it is not changing if C A is not changing if it is not changing then it is remaining constant therefore, we do not have to use that expression. So, what we are trying to put across to you is the following that in the case of a reaction in which thermodynamics affects the rate of chemical reaction. That means, this C A star is affected by the fact that K P is of magnitude that we must take into account then this effect will come in determination of this driving force C A minus of C A star is this clear. Now, having said this what is the value of K P what is the value of K P now K P from our equation here it is written as our K P is P C P D I will just write here K P is P C P D divided by P A. Now, we can write P C and P D in terms of concentrations and so on therefore, finally what you get here is that the X A star X A star is determined by this equation here. So, you determine X A star because K P is known C A 0 is known R T is known therefore, you can determine X A star. Once you know X A star once you know X A star we can also determine C A star we have done that already therefore, we are now in a position we are now in a position to we just put down once again. Now, since C A star is known in terms of X A C A is known in terms of X A therefore, you able to tell what is the rate at which the chemical reaction will occur because the effect of K P is now accounted is that clear. So, to cut this long story short what we are trying to say is that in this reaction carbon plus hydrogen steam giving you carbon monoxide plus hydrogen K P is only 2.6 therefore, the effect of K P must be taken into account. How did we do that we did that we did that by recognizing that K P is given by P C P D by P A I will put a star here saying it put a star here therefore, K P can be put in terms of conversion which we have done which we have done therefore, we find K P and X A star is related by this equation where once K P is known R T is known C A 0 is known therefore, X A star is known on other words we are able to tell what is the driving this I will have to divide here by 1 plus X A that effect may be this is not the way to do this. This is the at this point will stop by saying that K G C A minus of C A star C A star is known in terms of K P therefore, we are able to know what is this driving for C A minus of C A star and therefore, we are able to tell what is the effect of the equilibria on the rate of reaction. So, what we are trying to say here we will we will we will put this not like this we will put this as C A G minus of C A star this is the driving force not C A G. Now, what we have tried to accomplish here is that this time for complete consumption is previously given only in terms of C A G because C A star was not important. Now, the driving force is C A G minus of C A star and what is C A star and what is C A G both we can calculate from our understanding of thermodynamics what we have done here we just put it across to you once again. What we have said is the following what we have said is that C A and C A star in this case is given by these two equations C A and C A star is given by these two equations and if the if the composition is changing if the composition is changing if the composition not changing then C A G C A G remains the same it does not change. Now, if for some reason in the flow reactor in these compositions are changing you have taken to account the effect of C A G as the composition changes then we will have to use this expression C A star is given by this C A is given by this. So, that you can substitute for both C A G and C A star in this equation and therefore, you know what is the rate function. So, essentially what we are trying to put across to you now is that for a situation of a shrinking particle if the composition is changing along the length of the reactor we must take that into account by using this formulation that we have said here. If composition is not changing then of course, you do not have to worry about it, but both the cases are formulations are satisfactory because you have taken both situations into account.