 So, what you would like to do today is to look back at what we have done so far and then apply it for different applications. Specific example I would like to take today is oxygen sag in river systems. We have spoken about it at an earlier stage, but today we thought we look at you know how various numbers come together in trying to understand what happens to oxygen in a river and how the pollution load that we introduce into the river due to variety of reasons. Either man made or natural whatever that how it affects the rivers, oxygen supply and how we can restore things like that. This is what we would like to do in this exercise. So, let us quickly recall what we have done before. So, we have a river system going this is and you have let us say sewage enters the river. Now, this river may have a certain velocity and as this river flows and as this sewage gets mixed with this river and therefore, the sewage oxygen or what is called the oxygen demand oxygen demand of the waste. So, this oxygen demand of the waste will start to utilize whatever oxygen that is present in the water and whatever oxygen that diffuses from the atmosphere into the water. And therefore, you will find the oxygen availability would be different from what it was previously. We would like to of course, see how this oxygen levels deplete and how the oxygen levels restore itself and so on. So, let us write our to just to recap let us write our equations once again. So, we have we are writing a balance for oxygen say input of oxygen minus output of oxygen plus generation of oxygen equal to accumulation of oxygen. So, let us write it for steady state by steady state we mean that there is no change with respect to time as you go along the river. So, this is not there. So, let us say the velocity of the river velocity of the river is u. So, u and cross section area is v s a. So, at any position x minus of u a c at any position x plus delta x and then you have oxygen supply because of oxygen transfer. So, I will call that as some k 2 times c s minus of c some a delta x this is the amount of oxygen that is coming in from the atmosphere oxygen. And now you have the oxygen present in the river which is consumed by the pollution. So, let us say in due to our sewage input the sewage is whatever is in this region. So, you are putting the sewage. So, let us call the sewage load as s. So, this is the the oxygen consumed by the sewage plus you have of course, what is called as r r p minus of r s. So, these are this is photosynthetic oxygen production this is photos this is the respiratory oxygen demand and sediment oxygen demand times a delta x equal to 0. Now, this is something we have done I am just doing it once again just to recap what we have done before. So, you have the material flowing into this control volume let us say this is the control volume. So, much is coming in let me so much of oxygen is coming in so much of oxygen is going out. And then so much of oxygen is supplied during in this this is oxygen supply k 2 times you know multiplied by a delta x and then some amount of oxygen that is used up because of the pollution load which is this quantity. And then naturally you we have photosynthetic oxygen production you have oxygen consumption because of respiration the oxygen consumption in the sediments all that is taken into account and we call this as minus beta we call this as minus beta. So, let us just simplify this this please simplify this please help me so that we can put all these numbers. So, what I am going to do now is divide throughout by a delta x and take the limit as x tends to 0 and so on. So, when you do that you get minus u d c d x the first term can see here please see here divide throughout by a delta x. So, it becomes minus u a cancels of d c d x plus k 2 times c s minus of c minus of k 1 s minus of beta equal to 0 please make sure that our numbers our equations are proper you can see here. So, I have let me run through this once again. So, this gives you minus u times d c d x. So, this becomes no change in sign this is a minus sign and this is a minus sign nothing has been nothing nothing new has been done all right. Now, we have to solve these equations and we have done this before and we recognize here this s this s is s is pollution load pollution load and this pollution load we assume that it is d k s therefore, s at any time is some s naught e to the power of minus of k 1 times tau where tau equal to residence time residence time of pollution load load. Now, what we are trying to say is that if k 1 is a rate constant for the removal of the pollution that we put into the water and tau is the residence time therefore, any instant at any position after a time of residence you will find that the value of s would be this. So, that means s naught minus of s would have been the consumption of oxygen of the river. Now, we have done all these things if you recall we had defined a term called this is what is called oxygen sag this called oxygen sag. Why is it called oxygen sag you would generally expect the river to be at near saturation, but at the moment it is different from saturation. So, it is this difference which is called oxygen sag what is important for us is to recognize that the oxygen sag should be a small as possible showing that the oxygen level in the river must be close to saturation. So, that aquatic life is able to perform its natural functions. So, we have solved this problem we have solved this problem for the situations of our interest and I just write down the solution because we have done this before. So, we will not do this again because it is already well known to us. So, we will not do this again I just write down the solutions that we know. So, what we have shown before this is what we have shown before that psi equal to k 1 s naught divided by k 2 minus of k 1 e raised to the power of minus of k 1 tau e raised to the power of minus of k 2 tau plus beta divided by k 2 1 minus e raised to the minus of k 2 tau plus some this is the initial k 2 tau. So, this is the solution that we have already gotten now here k 1 k 1 is pollution removal removal rate constant and k 2 is re aeration rate constant. And psi i is initial sag what is initial sag it is C s minus whatever is the oxygen at the suppose we have this river like this. So, you have pollution coming in. So, as soon as you mix the two and whatever is the initial level of oxygen in the river as soon as it is gets mixed up with the pollution that is called initial sag called initial often it is called as initial deficit. Now, the problem that we would like to solve let me just quickly explain the problem that we want to solve. See the problem we want to solve is the following you have a river then you have it is coming in at thousand cubic meters per day. You are putting in 100 cubic meters per day of pollution containing C O D equal to 500 milligrams per liter. It does not seem to have much B O D and sense that this is an industrial pollution which is C O D B O D has been removed in the treatment what is left behind is only C O D. So, it is going into the river now it is coming with D O or oxygen as nil. That means, there is no oxygen in this water and while the river the oxygen is coming in at it is oxygen solubility is given as 6 milligrams per liter it is 90 percent saturated 90 percent saturation. So, it is here is C equal to 5.4 milligrams per liter is that clear. So, river velocity is 0.1 is not flowing velocity is low per river 0.1. Now, this these numbers are chosen just to sort of come as close as possible to some of the problems that we may encounter in tropical region such as ours. Because, particularly the decant rivers for example, there is in much water for much of the year excepting during the monsoons and whatever river flows it will be largely due to a dam which holds the water and then releases the water as per the requirements of the downstream population. Therefore, the river flows may not be very large. So, keeping that in mind the velocities mentioned here are very low. So, the question is let me just state the questions. So, you want to k 1 k 1 is given as 0.3 per day and k 2 which is rearation rate constant is given as 0.7 per day k 2 equal to 0.7 per day this is what is given. So, we have to find out max sag number 1 and number 2 where max sag occurs and 3 where downstream where downstream d o reaches 90 percent saturation and then where downstream and 4 what is the c o d when max sag occurs. Some usual questions these are all usual questions that we would like to know about a river when it is affected by human interference. So, some of these things of course, we have done. So, this is not new to us. So, let us look at this whole situation one by one and then try to. So, our equation please recall here that this is the equation that tells us what is the sag at for different residence times. Now, what is psi is defined as saturation minus of c i. So, first to start with we have to find out what is the initial state. So, let us find the initial state and then once you know the initial state we can do all the rest. So, what is the initial state the initial state is now let us say we have 6 into 0.9. So, this is 5.4 into 1000 cubic meter divided by total flow is 1000 plus 100. So, this is the d o in the river. So, is it correct. So, the multiplied by 100 multiplied by 0. So, what is the d o in the river now let us calculate this calculate what is the d o in the river as soon as the pollution reaches. So, we have calculate d o in river equal to 0.9 multiplied by 0.6 multiplied by 1000 this is what is coming in plus from sewage there is nothing is coming divided by 1000 plus 100. So, that you can do this simple calculation that comes out to 4.9 milligrams per liter is the oxygen. Now similarly, c o d in river c o d in river is how much you are coming up 500 multiplied by 1000 correct sorry c o d this means 500 multiplied by 100 100 is the sewage that is coming in and then you have 1000 cubic meters into it is coming in at what is mentioned let me just please check this what is coming at 5 milligrams per liter c o d of the incoming water is 5 milligrams per liter. So, divided by 1100 that is it is 1000 plus 100 is 1100. So, when you do this calculation it comes out to be 50 milligrams per liter. What are we saying now we have a river into which sewage has entered and then as a result of that we have c o d here is 50 milligrams per liter oxygen which is c i equal to 4.9. So, that is the state at the point where the sewage has entered and this is the state of the river at the time where we enter this. Our intention is what happens as it goes along how long what is the time it what is the residence time or what is the length of the river that it takes for it to restore itself and so on. This is the questions that you want to answer, but we have solved this kind of problems before. So, we can very quickly find out what is what is psi i by definition it is c s minus of c i which is 0.9 multiplied by 6 minus of 4.9 that becomes 0.5 milligrams per liter we start like this is that clear. So, our equation is here. So, you want to know when this how low this can become of course, we do some trial calculation etcetera. So, let us put psi equal to 0 and find out where this becomes at what residence time this psi becomes 0. So, when you put that when we put psi when you put c equal to 0 or psi equal to 5.4 milligrams per liter what is meant by that when you put c equal to 0. That means, at any position psi equal to c s minus of c correct. So, if you put c equal to 0 then psi becomes c s correct that is the highest sag that we can have. So, we can see at what position it this very poor very bad situation arises. So, when you put psi equal to that corresponding to 5.4 equal to what is the what is tau which means what we will have to put psi equal to 5.4 here you know k 1 you know k 2 you know s naught s naught is 50 you know all that beta we generally take beta as 0 because saying that basically this oxygen is not available for our purpose. So, we take beta as 0. So, only thing is that we have to take this term and this term. So, we can solve for given psi equal to 5.4 we can find what is tau it is a fairly not a difficult calculation I am not saying it simple we can do it graphically what is tau. So, we get solving I have done this for you solving we get tau equal to when you do this I get of tau about 0.4 per day 0.4 0.4 days is that clear what are we saying now what we saying is that that after putting this up to about after about 0.4 days our velocity is 1 kilometers per hour. So, which means that after distance of about 0.4 kilometers you will find that the oxygen has become 0 oxygen in the river has become 0. So, it is a point we are trying to say. So, which means that oxygen level in river becomes equal to 0 at tau equal to 0.4 days or distance d equal to 0.4 multiplied by 0.1 multiplied by 0.4 that is equal to 0.04 kilometers at this distance it has become very quickly it goes to very small value. So, this is the first question the second question is I mean it becomes 0 all right what happens to COD at this point what is the value COD at this point if I ask you what will you say. So, we can calculate that also. So, what is the COD at this point COD at this point is s equal to s naught e raised to the power of minus of k 1 times tau. So, s naught is 50. So, e raised to the power of exponential of k 1 is 0.3 with a minus sign multiplied by tau which is 0.04. So, this turns out to be that is about 44 milligrams per liter. Now, the point of interest to us is I mean what distance after which you know the systems start to improve itself. Now, in other words I mean we want this value 44 to become 5 because that is what the river was when it this way is entering. So, in other words what is tau when s equal to 5 milligrams per liter. So, how do we calculate that we put s as 5 and s naught as 50. So, we can calculate therefore, we can do the calculation. So, 5 equal to 50 exponential of minus of k 1 is 0.3 times tau. So, we can do this calculation 5 by 50 is 0.1 ln of 0.1 is 2.303. So, becomes 2.303 minus divided by 0.3 how much is this 2.303 divided by 0.3 equal to 7.67. So, tau equal to 7.67 days or implying the velocity is fairly low is 0.1 kilometers per hour. So, 7.67 multiplied by 0.1 multiplied by 24 equal to 18 kilometers. C O D level reaches the initial the earlier point at 18 kilometers. Small error here just correct this for you have to multiply this by 24 as I forgot. So, when I multiply this by 24 this becomes 24 is about 0.96 kilometers. So, let us just cut the long story short that. So, if I ask you what is the level of oxygen level let me write this here because it is taking too much space. So, if I ask you what is d o at tau equal to 7.67 days which means at after 18 kilometers. So, you can calculate that once again our equations are all with us. So, you have to once I say that tau is 7.67 you can find out what is the d o at that point. So, that means psi is known means d o is known because it is simply if you put psi c s is 6 or 5.4 you can find out what is d o. So, what we are trying to say here is that I mean that point is not to be forgotten that if you have an industrial ways coming into water. Our solution suggest that you know the river requires sufficient distance before it restores itself very clearly showing that you know our river systems really do not have the oxygen capacity to manage industrial pollution. So, we need to look at this problem seriously and see that you know we do not allow our waters to be interfered by you know COD kind of pollution. So, you must treat them prior to it. So, that what you put into the river is not COD of 150 and 300 and 500, but something like less than 5 less than so that you know the river has the ability to manage the pollution that is entry. So, what in other words you are trying to say that the natural aeration capacity of the river is really meant for the natural biology of the river and not for human generated pollution. So, human generated pollution must find its mechanisms of treatment and not be dependent on the river systems that is the point we are trying to put across. So, this one example is to illustrate how fundamentals of reaction engineering come together and to solve or to explain how we can manage our river systems and how we can keep it in good shape and ensure that our biology of the river is in good health and so that our dependence on the river systems for our health etcetera are ensured. Let us go to the next example. The next example we would like to look at is something which all of us have encountered that we want to see how we can use population balance modeling to understand a forest. We are still in the general area of environment see our forests are source of water, source of raw materials, source of food, source of fruits, source of timber you name it you know it is forest that really sustains human populations and unless we know how to look after our forest we will have problems in future and let us remember forest is not just you know greenery forest as animals and it is these animals and whose independence ensure that the forest is in good health. So, this particular exercise which you would like to do it read something like this it says we have a forest there are animals in this forest it says what does it say animals are born continuously at the rate of s minus of 0. So, s s is the age of the age of the animals, age of animals animals are born at a rate which is given and animals die at a constant rate animals die at a constant rate d which is given alpha times s where beta and alpha are constants. So, what happened you have you have a forest which is not there is no interference there is no there is there is no such movement out to the forest there is nothing coming in there is nothing going out. What we are saying is that this is it is a close system he is a close system in which animals are born animals grow animals die. Now, it is reasonably well known there is plenty of data that is exist around the world that a forest left to itself is at steady state by steady state what we mean is that if you take a sample of population wherever it may be that population does not change with time. On other words the birth and death process that is taking going on inside this environment called forest remains reasonably well balanced. Now, it is when only when we interfere that these things seem to go out of control. On other words of course, we all require forest for our dependence. So, we require to harvest various products of the forest as all this is well understood, but perhaps what is important is that whatever we interfere or whatever we take out of the forest must be. So, small compared to its internal regeneration capacity. So, that the health of the forest remains essentially unaffected and this situation of forest around the world is precarious as you all know whether it is Indonesian forest whether it is Amazonian forest or forest in the Indian some continent you name it all over the world it is in back shape and the more we can understand the fundamentals of this the better it is for us. So, this particular exercise is just to give you a feel for you know how things might happen just a model it does not mean that it is the best way to understand what is going on it is a very very simple way of understanding a very very complex phenomena that might be in front of us. So, what is it that we want to do what we want to do now is to write our number balance about what is happening in this forest. So, we have let us say let us out the balance is some input output generation equal to accumulation. This is our balance that we have been writing for a long time we must adapt this for our forest let us adapt this now. So, I will still carry on with our forest. So, input output plus generation or birth minus death both are there equal to accumulation. Now, we say that it does not change this is the most important point that we want to recognize understand and carry home in our message that forest is essentially at steady state and if we must understand the fundamentals of how it keeps itself at steady state. So, that when we intervene if at all we intervene we would like to intervene minimally. So, that this is not disturbed too much after all we require the forest for our timber we require the forest for our for our fruits you know we require it for our firewood may be and we require it for you know managing the carbon dioxide oxygen balance. So, that you know the in the aerobic processes of this environment and the photosynthetic processes are all in balance. So, many issues are connected with a healthy forest. So, we are only looking at a very small aspects of what is going on. So, let us say our balances if you recall when we wrote our balance at an earlier stage it looked like this. So, that is when we wrote it for a process continuous where it is coming in and going out this is f naught this is f 1. So, this is the balance that we have been writing for a long time if you recall. Now, we want to write this for the case of a batch process that is what we want to write this for. So, let us learn to write this let us learn to write since there are no inflows and outflows. So, these terms must be deleted these terms must be deleted we know that, but we must add to this because add to this the birth and death terms the birth term which is beta times delta of s minus of 0 and then you have the other term which is alpha s. So, this is the birth term this is the death term correct equal to 0. So, what is it that we have done what we have done is that we have written the material balance we have derived all this therefore, I am not going through this once again we have derived all this. So, we have the population balance in which we have deleted the input term we have deleted the output term because there is there is no input there is no input there is no output in the sense that there is no flow out of out of the forest we have removed those two terms. So, this is the term which is what we call as the change in age and so on and this is the birth term this is the death term. So, what do we have beta beta delta of s s minus 0 I have not written minus alpha of s equal to d by d s of f 1 r 1 b is it alright what we are saying. Now, when we talked about this is what this is rate of change of property, what is our property this property is age. So, it is in a sense it is time. So, our this r 1 by definition is what I will call this property as s. So, we have d by d t of our property which is s which is age and age is time. So, here s equal to t and therefore, it is equal to 1 something that we have said when we talked about residence time distribution in a stirred tank if you recall we did say this at an earlier stage when we talked about residence time distribution in a stirred tank and there we said that. So, there we said that if you want to find out the residence time distribution why I am doing to do this again and again is that if you learn how to apply population balance where we are talking about numbers then we can deal with variety of situations of daily life this might be what movement of shares in a market this might be you know how you know population of a town village etcetera changes variety of things that we would like to know in daily life. So, I am just thought it is important that we try to understand this that it is a closed system there is no input there is no output. Therefore, remove the first two terms generation is because of birth and then death there is a continuous it can happen at all time. So, we have taken both these terms and therefore, this particular term which talks about what happens inside the forest that effect is also taken into account. So, birth and death how they affect how the population changes that effects are shown here. So, we have our equation I will put r 1 equal to 1 here minus minus equal to 0. So, our equation looks like this that beta times delta of s minus of 0 minus of alpha s minus d by d s of r r r 1 equal to 1 we have shown just now multiplied by v multiplied by your distribution function f 1 this is is it clear r is 1. So, we have our equation beta of s minus of alpha of s equal to d by d s of v times f 1 beta by v and alpha by v you know they are constants beta by v alpha v are constants you know they are constants beta by v alpha v are constants. So, that now we have I can I mean for the moment I denote beta by v as beta alpha by v as alpha for our for the sake of this one. So, I will retain this as beta maybe I will can call this as beta dash. So, that you know this becomes beta by v is beta dash by v is beta alpha dash by v is alpha. So, it becomes alpha s equal to d by d s of f 1. So, minus minus all minus equal to 0 equal to 0 now. So, delta of s delta of s delta of s delta of s. So, our equation. So, what is it that we have now what we have now is beta times delta of s minus 0 minus of alpha times s minus of d by d s of f 1 equal to 0 this is the equation that we have to solve is it. Now, just want to spend a few minutes here and go through the whole thing once again please bear with me I will put a dash here. So, we have a forest where the birth rate is given as beta delta s minus 0 death rate is given as alpha dash s minus of 0 this is our equation that describe the number density. Now, we said that you know this input output this how we started our population balance, but there is no input there is no output that is how this equation looks like this birth and death and so on. So, then we said that we can write this as beta dash and then so that when you put this rate of rate of change of property r 1 as 1 because we said s refers to age really s refers to time basically therefore, d by d t is equal to 1 that is how we substituted r 1 as 1 then we said r 1 is 1 then beta dash by v we said beta and therefore, we got this equation therefore, we have this equation which will tell us this is the equation which will tell us how the what is f 1 it is the age distribution age distribution in forest how the age distribution in forest and how this age distribution is affected by beta and alpha. On other words how birth and death process adjust themselves to keep the age distribution unchanged in a forest this is what we are trying to say. Now, to solve this equation we said that we will remove the unbounded term and solve the homogenous equation and then to get the constant of integration will generate a boundary condition. Therefore, solve this to first solve homogenous equation what is the homogenous equation d f 1 by d s minus of s equal to 0. So, what is the solution what is the solution we can say f of s equal to minus of alpha s square by 2 plus c it is a very elementary thing you will there is nothing to there is nothing much here I called f 1. So, solution is f 1 of s is alpha s square by 2 plus c this is the solution. So, how do you find constant of integration how do you find constant of integration. So, to find constant of integration to find to find constant of integration for that what we do is that we take balance take balance between s minus of d s and s as s tends to 0 d s tends to 0. So, we write all the inputs all the outputs all the generations and then birth death and so on equal to 0. So, there is no input there is no output. So, what is the generation we said f 1 r 1 v at s minus d s f 1 r 1 v at s minus d s f 1 r 1 v at s. Then you have generation which is beta delta s minus 0 s minus of 0 d s beta s into v minus alpha s d s v equal to 0. This is all right with everybody what we have said we are writing we are writing balance between s minus d s and s. Therefore, this is material generated at s contributing to s minus s minus d s to s and then s to outside. Therefore, this is the difference that accumulates in the interval this is due to birth and this is due to death and there is no generation this is clear what we are saying. Now, we have to do a limit at s tends to 0 d s tends to 0 d s tends to 0 this goes away there is no element that belongs to less than 0 it goes away and then beta delta s d s this whole thing becomes. Therefore, this from here we get f 1 r 1 is 1 v cancels off can see here f 1 at 0 equal to beta. Is that clear what we are saying see how beautifully the material balances come together. So, what we have said here I mean this is common sense this is also common sense. So, what we got we have got here please notice. So, our solution. So, our f 1 of s we said is minus of alpha s square by 2 plus constant and that constant of integration and what we have said f of 0 equal to beta. So, you put here beta here this gives you c equal to beta. So, our solution is equal to minus of r square by 2 plus beta. Now, we should have integral f 1 d s equal to 1. That means of all ages all ages when we integrate over all ages this equal to 1 by definition. Therefore, that implies let me just do this once again. So, f 1 of s equal to minus of alpha s square by 2 plus beta. Then integral if it takes integrating please help me integrating over all ages all ages assuming s 0 to 1 normal as all I will call it normalized age you know when I say normalized age. So, this becomes 1 equal to minus of alpha s cube by 6 plus beta this is all right beta shall we say beta is that correct beta s going from 0 to 1. So, this becomes 1 equal to beta minus alpha by 6 or we get beta equal to 6 minus of alpha by 6 or we get beta equal to 6 minus of alpha by 6. So, this is the kind of relationship that exist between alpha s and beta s. Let us just quickly review this what we have said what we have said is that if you have a forest if you have a forest and then there is birth and death process in the forest. Then you will I mean in order that steady state is established we must have that the birth and death process are just themselves. Otherwise the forest will not be a steady state and this is. So, if there is no external interference there is an interference clearly these problems are for more serious. So, the point of this exercise is that we must appreciate and we must design our approach. Our inroads into forest with an appreciation of the fact that alpha and beta are related and we must not disturb that relationships. That is the fundamental message that this kind of problems that bring it to us if to our daily life. So, let us go to the third exercise a relatively simpler kind of exercise which we want to look at now which is what we call as the third exercise we want to look at of course, it is a very big industry steel making is a huge industry in this world and steel making comes from blast furnace as you all know. Now, then you must start asking you know why is it that sponge iron which is a more recent kind of technology and why is it is gaining more and more popularity the reasons are many. Now, the reasons are that blast furnace technology is suitable for very very large scale productions. Now, and therefore, huge investments and then huge interference in the environment and so on. But, sponge iron is more expensive technology, but it is suitable for much smaller scale operation. See that is the most important part of sponge iron technology and I thought we should spend some time and more importantly that you know should appreciate some of the chemical reactions that are so central to the sponge iron technology. So, that we appreciate how you can also make sponge iron for various purposes because once you make sponge iron it is then used for steel making and this is the next part of the process. So, we are looking at sponge iron what is the technology is Fe 2 let us say Fe 2 O 3 plus thrice H 2 use it twice Fe plus 3 H 2 O. Now, notice this K p for this reaction at 25 C is very small. The heat of reaction heat of reaction turns out to be 7 kilo cal per mole it is endothermic you need to supply heat energy to be able to conduct this reaction. Now, let us just look at so you have a solid this is you have hydrogen diffusing as hydrogen goes inside goes inside and then it starts to react and then we say that you know the core keeps on keeps on drifting inside. So, it is a shrinking core we have done that shrinking core. So, if you look at a model is assuming that it is a spherical particle. So, as reaction proceeds we would expect that the this is the reacted layer this is the unreacted layer. So, some other time if you look at another time. So, may be it is like this this is the reacted layer. So, what we have we have external diffusion external diffusion of hydrogen what is meant by external diffusion that means there is a layer through which the diffusion of hydrogen occurs. So, there could be a resistance to external diffusion then there is internal diffusion this could there could be a resistance because of internal diffusion which is the product layer. And then there could be a what is called resistance due to chemical reaction that means the chemical reaction also there is certain amount of resistance. So, essentially in any of these reactions you should expect that external diffusion internal diffusion and chemical reaction all three of them could be important. On other words we must be able to calculate what is the resistance due to each of these controlling steps. Let us quickly calculate these controlling steps. So, that we have an idea of the the resistances if we do a quick calculation a quick calculation for you one minute and just write the equations down and then. Now, we have if you recall if you recall that we have shown we have shown if you have a single pellet. See we have done this we have done this before we have taken spherical pellets and then we have shown that the time for certain extent of reaction is given by r c by r. Let me explain this give me a minute and then I will be back with you r c cube by r cube plus 1 minus thrice r c by r whole squared twice r c by r whole cube where we know that r c by r equal to r c by i equal to 1 minus of x b to the power of 1 by 3. So, I am just saying a gas plus let us say b b solid equal to product types. So, if you have this kind of reaction. So, we have the time that is required for the reaction. Suppose r c by r can be put in terms of x b therefore, let us say we want 50 percent conversion. Therefore, if you replace r c by r like this we can calculate what is the time that is required for 50 percent conversion 60 percent conversion 80 percent conversion and so on. On other words what we are trying to say here is that we can calculate the time for given extent of reaction. What is all these numbers tau r let me just put them down because this is I will put them in the next page because not enough space here bear with me I will just write here very. So, that you know t equal to tau r 1 minus of r c by r tau f 1 minus of r c cube by r cube plus tau d. 1 minus of thrice r c square by r square plus twice r c cube by r cube. Now, tau r we know this rho b r divided by b times k s times c a g rho f we know all these things from our previous exercises thrice b k g times c a g and tau d. Is equal to rho b r square divided by 6 b d c a g what we are trying to say here now is that since these are all known quantities because we know the concentration we know the rate constants will tell you how to calculate all this in this particular example. So, what we are trying to say is that if you have sponge ion to be to be manufactured then if you know all these numbers for this reaction which we will do shortly then we can calculate the time that is required for this reaction to go to a different extent. So, we will look at this in greater detail we will meet next time. So, what I am trying to put across to you is that sponge ion technology essentially is an example of reaction between ferric oxide or it is called hematite with hydrogen to give you ion and this is what goes to steel making this goes to steel making. Thank you very much.