 Today, we will be looking at this biomethanation reaction. Now, what happens in biomethanation is something like this. Generally, you have insoluble organics. Say if you are going to a let us say a dairy industry for example, the waste organics lot of solids are coming. So, insoluble organics it gets digested by extra cellular enzymes in the reaction equipment and becomes soluble organics. And then volatile acids forming bacteria works on it to give you acetic acid. And then gasification by methane producing bacteria to give you methane and carbon dioxide. So, it is a 3 step reaction that is what you see in the biomethanation process. Stage 1 is solubilization, stage 2 is acid formation, stage 3 is gasification using methane bacteria. Now, generally we find that it is not a bad assumption to assume that the methane producing step is the slowest step of the reaction. So, the most of the designs therefore, is done on the basis of this assumption that methane producing is the slowest step. Although there are instances where this solubilization is quite slow as a result we may have to take this also into account. So, as far as this problem is concerned we are looking at production of methane using acetic acid as a substrate and that is the chemical reaction that is set up acetic acid. So, this is the chemical representation of what is going on acetic acid this is the nitrogen that is present in the environment to give you some bacterial growth which is this is the chemical representation of bacteria and methane and carbon dioxide. So, you notice here that roughly equal quantities of methane and carbon dioxide is produced volume basis. There is lot of literature in bio methaneation over the last 30, 40 years many, many plants have been set up all over the world. It is a reasonably well known technology there are many problems with it we will talk about it as we go along. Now, what seems to be an appropriate growth function is let me write down the growth function. The growth function is something that we have set up it looks something like this mu m mu equal to mu m within brackets 1 by 1 plus k s by h s and h s by k i. This was proposed by Greif and Andrews way back in 1974. There is lot of material on this what is h s h s is the anionized acid. So, this is the anionized part of acetic acid which is what is important as far as the growth function is concerned. The numbers are given k s is given k i is given this form of rate function. We have seen earlier also this is a form where substrate inhibits the rate of chemical reaction. Therefore, we would like to operate the process at h s value equal to k s times k i square root. So, the best choice of h s which is the substrate which is active as far as the process is concerned is square root of h s times k i. You can calculate that from here. So, what you want to look at is a very important area dairy wastewater. I have just taken 3000 cubic meters per day 6000 milligrams per liter 6000 roughly 100 millimoles not mole 100 millimoles 100 millimoles per liter is the concentration of acetic acid that will come out of this 3000 milligrams per liter. So, 6000 milligrams per liter it is being treated in a chemostat. Chemostat is our CSTR equivalent of which we have talked about already. What you want to do is what just get an idea of the size and what are the outputs that we can expect. Now, when you go around the world you find that dairy wastewater what they would do is that they would first do a biomethanation and as you will see as you go around this problem we will see that biomethanation is not able to completely remove or consume this 6000 milligrams per liter. It is only able to do part of it. On other words the effluents that come out of dairy wastewater still is very concentrated from the point of view of discharge. So, you will have to do another step of treatment before it is suitable for discharge. So, we will have to look at that also to find out how to handle this kinds of problems. So, if this is the rate function where mu is mu m times this one. So, what is the best choice of h s that we will take? What is that value tell me? What is that? Please calculate and tell me this is the ionization reaction here k a is given k a is given somewhere I think k a is 4.5 and operating p h is 6.8. So, you can find out what is the h s value? So, the h s at which we will operate is 0.15. Is that ok with everybody? 0.15 root of k 1 k s k i square root is 0.15 is it alright? Do we all agree? This is what I have got h s optimum I have got 0.15. So, as per the data given the best choice mu equal to d correct we must have mu equal to d chemostat mu equal to d. Therefore, the dilution rate at which we will operate is 0.27 per day. Is it ok? Can you please calculate and tell me whether this mistake in my calculation? 0.27 per day is ok, 0.27 is alright. So, what should we have? So, our equipment volume should be 3000 divided by 0.27 that is about 11000 cubic meters is what you get now. 3000 divided by 0.27 is about 11000 is it alright? 11000 is ok. So, what is the now if it is 11000 what is the value of s minus I will get is a 1795 s minus comes from the acetic acid equilibria here. What is the equilibria is given? All the numbers are given yes or no? So, what is the acetic acid that you have started with? Still remains unreacted the acetic acid that still remains unreacted 1795 is s minus how much is h s? Can you calculate what is h s? At the end of the process s minus what is the h s? Is this correct? If there is a mistake tell me I will correct it 0.15 millimole per liter 0.15 multiplied by 60 is 9 is it alright? See we are operating the process as optimum value of h s which is optimum is 0.15 millimole per liter that is what you told me multiplied by 60 is the molecular weight of acetic acid. So, I put it as 9 is it ok? So, what are we saying that we are the process we are feeding at 6000 and the biomethanation reduces it to 2000 is that clear? So, biomethanation process is able to reduce the organics to a level of 2000 it is not able to reduce it to a level that is suitable for discharge. So, it is 18 and sorry. So, I have to make some adjustment in these numbers. So, these numbers are not exactly correct ok please help me now what is the loss of COD s naught minus of s? So, this is it is not exactly 6. So, you tell me the actual numbers please. So, it is 6 minus 1.8 not 2. So, this is not exactly 12000 it will be slightly less slightly more. So, please tell me these numbers 1 2 1 2 6 12600 is it ok? Is it alright? So, what are we saying that so much of loss of COD s happened 12600. Now, if you look back at our stoichiometry it says what does it say CO 2 and they are roughly in equal volumes. So, in terms of mass basis I have calculated here. So, based on this just look back at this one for every 60 we get 0.92 times methane which is 0.92 times 16 and 0.92 times 44 correct. So, that is y axis this is y y p s and this y p carbon dioxide. So, that is what I have done. So, fraction methane to substrate is 0.24 I get carbon dioxide substrate is 0.67 please verify if it is ok 1 6 how did I get 1 7 9 5. So, brilliant question see 29.9 multiplied by 60 is it ok alright is it ok with everybody. Now, please tell me this numbers are correct 0.92 times 16 is 0.24 and then 0.9 to 44 is 0.67. So, if I ask you what is the methane fraction in biogas what will you tell me typically weight by weight methane fraction 0.24 divided by 0.91 whatever that number is. So, you will find in biogas on weight basis the methane is not more than about 26 27 percent ok on volume basis it is equal on weight basis it is only 25 26 percent is it ok with everybody 0.24 0.67 alright. So, what is the methane production I have made a small mistake here. So, you can tell me. So, it is 600. So, this number is how much please tell me 3000 25 0.26 is it just a minute there is some important questions are coming please understand. What is the mistake we have made here please tell me COD losses 12600 that is this is the acetic acid loss out of that we said this is the fraction that is why you multiplied by 0.24. What we are saying is for every 60 we are getting 0.92 times 16. So, that is the fraction we are talking about is it ok is it ok please everybody what we are saying alright. So, where are we. So, we have 3000 kg of methane and. So, many kilograms of some slight mistake is here what is this number please tell me 12600 840 ok fine. Suppose I ask you what is the thermal value of biogas what will you tell me what is the thermal value of biogas. So, the methane I have calculated here just to put it in the context methane is about 14000 kilo calories per kilogram taken this from the literature. So, biogas is about close to 5000 kilo kelv 3600 per kg or per cubic meter you can also calculate that I have done that. So, 3640 is what we get showing that biogas is a lean gas it is not a very rich gas it is a lean gas. And therefore, it is combustion in a IC engine will require suitable design which people have done already is very popular around the world biogas engines are operating very well. And what is the problem that we face biogas in IC engines the problem with IC engines is that in this reaction if there is protein generally you produce hydrogen sulfide here. And that creates problems in the combustion for which in small scales it is not easy to handle in large scales you can remove the hydrogen sulfide ok. What we find is that biogas yes it gives you very good thermal value, but it produces a waste water which is not ready for discharge ok. So, we must look at doing a system design in which the waste water is taken care. And this waste water we said the effluent that comes out of biogas this is 1804 somebody said. So, we said yesterday that we are systems will have a dilution rate of this type correct where mu m I have taken whatever you have done yesterday. So, this is the kind of number I get. So, the second system which will take care of the waste water is that it has a dilution rate of 0.5 and the system volume of 6000 is this ok please tell me this is all right 3000 cubic meters per day is coming the dilution rate is 1 by b times mu m s k s and all that I have taken of yesterday values or given yesterday. So, d value is 0.3 this becomes 0.5 and this is what I get please tell me if it is ok is it ok. I have taken b as 0.3 mu m as 0.3 per day k s as 10 this is in problem number 1 same thing we have done this yesterday same number say b is 0.3 everything is what we have done yesterday ok s and o shall we go forward all right. So, we have an equipment size of 6000 cubic meters which is for treatment of that waste water and what is the size of the equipment of the biomethanation what is biomethanation size we got 11000. So, you can say 11000 plus 6000 17000 cubic meters have been used up. Now, let us look at the real problem the real problem is here see what happens is that is this correct total COD loss that in this in bi this is in biogas is this correct what I have written 2000 it is 2000 see we have to remove see we have to see what comes out of the biogas plant is at 2000 milligrams per liter 3000 cubic meters. So, the total amount of COD that we have to treat in the effluent treatment plant or the ETP they call is this is this correct. Let me repeat what comes out of the biogas plant is 18 I say 1804 1.804. So, this is slightly less. So, it is 540 kg per day. So, what comes out of the biogas plant comes at 1804 I have written 1.8 kg per cubic meter 3000 cubic meters. So, 5400 kg per day has to be handled in your ETP plant correct this is what we are saying I am coming to that I am coming to that. Now, how much oxygen is required to treat to oxidize 5400 kg of oxygen demand if it is typically glucose 1 is to 1 is reasonable number. So, to be able to process 5400 kg. So, you will require 5400 kg of oxygen. Now, you can ask how this number comes from of course, this number comes from practice if you go to any waste treatment plant they will tell you that every kilogram of oxygen demand you roughly require 1 kilo kilowatt hour of energy. That means, your aeration equipment the amount of energy that you must put in. So, that you know it is sort of bubbles through this is about 1 kilowatt hour for every kilogram of oxygen you have to supply is this clear. So, if you have 5400 kilo I mean kilograms of oxygen to be supplied you will require 5400 kilowatt hour of electricity. So, how many kilograms of methane did we produce 3024. So, to produce see what we do in a this one is that we burn we burn methane and then converted to electricity. So, your 3024 kg methane burning and then generally efficiency is between 0.3 to 0.2 in that range. On other words 20 to 30 percent typically 22 to 23 percent of the thermal energy is converted to electricity. So, I have just estimated this numbers are not correct numbers are not correct please tell me now 3024 14000 by 860 what is the number correct if it is 0.3 what is it if it is 0.2 what is it 14 I write 770 if it is 0.2. So, what we are saying is that if you burn this you can get. So, much of electricity if it is 30 percent efficiency. So, much if it is 20 percent efficiency. So, much generally you will find that I C engines efficiency is a quite good. So, you might get something in this range say again by 860 see kilo calories to kilowatt hour conversion see 860 kilo calories equal to 1 kilowatt hour this is something that I have learnt in my school. So, what we are trying to say here is that see in ETP you consume. So, much and you generate. So, much. So, there is a slight power surplus this power surplus which is used for various other purposes is this clear. So, when you have biogas plant you have some surplus energy which can be used for various other purposes. Part of the reason why biogas is a very popular in various places is that it does give you some power surplus very good question that is just go back to that question where are we we calculated C O D loss for the biogas plant which is converted to biogas. We calculated C O D loss in the ETP plant which is converted to carbon dioxide. In both cases there is a loss of carbon in one case the carbon is given comes to you as methane in another case the carbon comes to you as carbon dioxide is that clear. Now, the important point to recognize here is that in many places you will find when they produce power production using gas the wasted energy which is the exhaust from the engine. In many of this dairy they need lot of hot water because dairy requires lot of washing to be done in various places. Therefore, they are able to use that energy to generate hot water and because of that the thermal efficiency of the process because they are able to recover lot of heat from the waste heat thermal efficiency is a very high. On other words many of these dairies if they are able to integrate biogas waste water treatment and therefore, that waste water they are able to convert it into hot water and use it for various kinds of washing. So, it integrates very nicely in terms of thermal in terms of say protecting the environment and so on. So, it is very popular around the world if you can afford the investment on the biogas plant investment on the waste treatment both are very large energy intensive investment intensive activity. The second exercise alcohol fermentations are status in this country of course, all the alcohol in this country is produced from sugarcane molasses and in many parts of the world particularly the warmer regions of the world it is from sugarcane molasses. Now, this data that I have taken is a rate at which the substrate gets consumed is given by this function mu m s k minus k p and all that these numbers by enlarge hold for sugarcane molasses and these numbers 0.17 all these effects are quite. Now, what happens in a sugar industry is that the molasses comes to you as a 50 percent sugar 44 45 sometimes 48 percent 48 percent means 400 to 500 grams per liter as hexo sugar. Now, industry what they will do is that they will dilute this to about 10 percent they will dilute this from 500 or 455 450 grams per liter to 100 grams per liter the question is fermentation alcohol industries typically dilute molasses 200 grams per liter y. Now, you will realize that let me write this stoichiometry here c 6 h 12 o 6 giving you twice c 2 h 5 o h plus twice c o 2 this is 180 this is 92 this is 88 on other words strictly anaerobic conversion of hexo sugar to alcohol is actually roughly 49 51 percent roughly say 50 50 is that clear. Now, if you start with 1 gram of sugar you will get half a gram of alcohol. Now, if it is if you start this as 100 grams per liter then you should get 50 grams per liter of alcohol correct is that clear. If you had started with 500 grams per liter you would have got 250 grams per liter of alcohol. Therefore, the amount of water you have to remove in this case it is only 4 times you know 250 to you have to only remove for every k j of alcohol you have to remove 250 k I mean 750 k j of water here you will have to remove 950 k j of water just see the difference. If you could have used 500 grams per liter sugar you are removing 750 k j of water for every k j of alcohol here you are using for every 50 or if you are removing 950. So, 1 to 19 here it is 1 to 3 you can see the advantage if you can use a higher concentration is that clear alcohol industries they do not use 500 grams per liter they use 100 grams per liter. They are willing to sacrifice this advantage why because the energy cost of distillations are very high we all know that, but in spite of the fact that the energy cost is. So, high they dilute it to 100 grams per liter and not work with 500 grams per liter. So, question is why is the question clear to all of you why do we dilute molasses I am say it again molasses is available in a sugar industry typically 500 grams per liter as hexo sugar. Now, this is diluted to 100 grams per liter hexo sugar prior to fermentation question is why what I am trying to explain here is that if they could have worked with 500 grams per liter they would have got 250 grams per liter of alcohol. Therefore, for every 250 grams they have to evaporate 750 grams of water 1 to 3 for every k j 3 k j of water has to be evaporated, but if they do this they have to remove 50 to 950. So, 1 to 19 the amount of water to be removed in this case is for every k j 19 k j of water has to be removed while here it is only 3 k j. So, there is such a great advantage in distillation when you can use 500, but they only use 100 the question is why the answer is there are many answer many reasons why they do this first reason is that molasses come is extremely viscous it is so viscous it is not easy to pump at all if you want to pump molasses you will have to dilute that is the first reason. Secondly molasses contains lost of suspended solids which comes from the process see what we do is that we have crystallize the sugar. So, we are just rejecting what we are not able to recover correct. So, it has lot of suspended solids and to be able to remove the suspended solids you have to reduce the viscosity. So, that is another reason why they dilute the third reason they dilute is this that this negative effect of alcohol accumulation in the process which inhibits the rate of chemical reaction what is called product inhibition all these factors combined make it essential for them to dilute it to a level which they can work with that clear that is why they dilute. Third is this reason that alcohol accumulation in the process inhibits the rate of chemical reaction because of this minus k p can look at this reason this minus k p effect. So, that makes it worse so that is why they would dilute thank you my friend. Let us go forward now now let us quickly calculate let us say we have the typical size of alcohol discleries in Brazil is very big, but ours is only 40 tons per day is typical and the biggest would be a little larger. So, if you are looking at a 40 tons per day alcohol factory. So, I just made some calculation a 40 tons per day alcohol factory how much should be this f 40 tons per day is this correct this see whether what is this calculation is correct 40,000 kg per day I have divided by 24 and then this is the s naught is 100 s is 10. So, we are taking it out a 90 percent conversion. So, I put is 90 and you can see here y p is 0.4 only 40 percent of alcohol is we get 48 percent is carbon dioxide is it all right and the remaining 40 and then this is 48 what happens is remaining this is 8 it is going elsewhere. So, only 8 percent is cells 96 till 4 percent is unaccounted there are other products that are formed in fermentation you know that little bit of glycerol and other products are formed that is why the data is not there. So, what is f equal to 46 is this correct 46 46 cubic meters per hour. So, what is the concentration of alcohol in solution what is the concentration of alcohol in solution 36 kg per cubic meter it is 3.6 percent 3.6 percent is the alcohol in solution. So, on other words if you start with 1 kg you get 36 percent is alcohol the rest is going to carbon dioxide see you can see here 8 percent is cell 48 percent is carbon dioxide and there are some side reactions like glycerol and all that which takes away little bit. So, what the useful alcohol is only 40 percent not much more than that and please recognize 90 percent conversion of hexo sugar is also not very common you know the reason is this inhibition makes it quite difficult to handle as alcohol accumulates this inhibition becomes worse. So, you will find a process a very good factory in India would be 82 83 percent not much more than that. So, we are looking at is here we are getting 36 per 36 kg per cubic meter, but in actuality we are not looking at much more than about 30 29 30 like that that is what is the common. So, the amount of energy that is required to recover this alcohol is very large. So, for 36 means you have to remove 9 960 kg of water has to be removed to be able to concentrate this. So, this is a serious problem the problem with alcohol fermentation lies really in the chemical reaction equipment we have not been able to figure out what is the way to take it forward. Now, an interesting work got done in Berkeley 1970 the very old research they spent a lot of time in trying to see whether this this problem that we are facing here of very lean concentration of alcohol and very high energy cost of recovery can we do something about it 1970s 1980s it still goes on and then various kinds of answers have been given that we quickly run through that because it is the context is important, but before that let us finish of this what is the dilution rate at which we must operate the plant what is the dilution rate mu m mu m is 0.5 s is s is 10 is it right s is 10 s n o s is 10 and k is minus 0.07 p is 30 things 0.22 is correct 5 25 roughly seems all right. So, what is the size of the equipment suppose you are producing 40 tons per day of alcohol what is the size of the equipment that you will have size of the equipment what I have got here about 200 cubic meters per hour now this is the term see what is the productivity for alcohol d times p is about 8 kg per cubic meter per hour or 8 grams per litre per hour is it clear what we are saying and this is what makes the whole technology little unviable because per unit volume of equipment you are not producing very much in 7.9 kg per cubic meter per hour. Now, if you look at the reality of alcohol production in India what you find is situation is much worse because this is for a continuous process where there is no shut down time you know filling and emptying which is a substantial amount of time is required to fill and remove and then in between also you have to clean the equipment and so on you will find because of the fact that the see here our residence time is about 5 hours correct. But you will find there the actual filling time and then cleaning time etcetera it becomes 18 hours. So, this what we are getting as 7.9 actually more like 2 or 3 grams per litre per hour is what we achieve in a commercial process is this point clear to all of you and all the problem arises from the fact that we have not got the chemical reaction technology right. The real we do that engineering of the reaction is where the problem is that is why all these problems are coming this therefore, just look at what was done. So, people have looked at what can we do I mean from the point of view of reactor design is there anything we can do is that clear is the problem clear to all of you. So, you have just done a calculation for a continuous process we said that it will productivity 7.9 correct based on the fact that we have produced 36 kg per cubic meter and our the dilution rate is 0.22 and therefore, our productivity 7.9. If you had run a batch process then the actual residence time would have been not 5 hours more like 18 hours 20 hours sometimes even more. So, you are getting much much lower productivity in the batch equipments that you see operating in this country. Continuous fermentations are not very common around the world not very common there are only few factories. If I ask you why is it that continuous fermentations are not very common people prefer a batch what would be our answer why is it that people prefer to do alcoholic fermentation in a batch why would they prefer batch in fact you would see if you go to places like Madhya Pradesh and Uttar Pradesh and all that there are much smaller distillates 5 kilo liters per day 5 2 kilo liters per day small plants are running you see and therefore, the productivity is a much much lower and therefore, many of these alcohol they are not going for pharmaceutical not controlled by government where is issues are there you know in the commercial sector, but the problem is that this productivity is a very very low and there are not good answers yet. So, let us just review what has been happening in this fermentation technology alternatives for alcohol a very nice book I mean it is a PhD thesis 1971 Berkeley it is in our library it is only about 120 130 pages and they have just looked at batch chemostats immobilized cell reactor vacuum fermentation and some combinations also they have looked at to see whether we can get around this problem of very poor productivity of the reaction equipment. So, what they have done is that see this batch reactor correct now if we have to overcome this problem what do what can we do see this inhibition due to product if you want to knock out this problem what can we do productivity is low, but we want to somehow get rid of this alcohol you have to remove the product how do you remove a product. So, they said let us do vacuum. So, what happens if you do a fermentation and the vacuum. So, you have an equipment which is operating at something like 30 millimeters 30 35 millimeters of mercury what would happen to alcohol it will go into the vapor phase alcohol will go into the vapor phase and water will also go into the vapor phase not just alcohol water would also go into the vapor phase. So, what will be the net effect of that as far as product in the in the broth or in solution concentration will come down correct. On other words as the fermentation proceeds alcohol in the broth would go into the vapor phase we continuously remove it because it is under vacuum therefore, the reaction keeps going forward. So, continuous fermentation essentially helps you to knock out this effect. So, that reaction is able to proceed in the forward direction and give you a much higher level of substrate conversion. This has been proven in the laboratory for 1980s lot of research is there which shows it is a very interesting technology from the point of view of driving the sugar to it is nearly complete conversion, but a huge problem came in try to commercialize this. So, what happened was this gas phase which contains alcohol and water which is at a pressure of 30 millimeters of mercury if it has to be you know taken to the next stage of the process you have to compress it. So, you have to compress from 30 mm to atmospheric pressure. So, the then it can be released into the atmosphere into a tank which contains alcohol. So, you have to compress from 30 mm to 1 atmosphere that compression is a huge cost. You have spent money in creating vacuum vacuum also cost money correct there is a energy cost in vacuum you need power to compress the products to atmospheric pressure. So, the energy cost of production became very large. Therefore, while you know the technology of it seemed very attractive you know drive the reaction to completion you see and therefore, since you are able to drive the reaction to completion you can use much higher concentration of s. See the your s need not be 100 dams per liter it can be 500 dams per liter it can be anything there is no great problem. So, you have an advantage of very highly concentrated alcohol I mean it is from 3 percent you go to something like 16, 17 percent because the lot of water vapor in the gas phase. So, you do not get very high concentration 20 percent you will get. So, to that extend you can save on energy cost of distillation, but the net result was that the energy cost of vacuum operation energy cost of compression was so large that the benefits of using a much higher s was not satisfactory. So, after may be about 10, 15 years of lot of research it had to be given up in fact, the last I mean chapter of this thesis I mean sort of says very surprisingly it says that may be 5000 years ago they understood this that is what it says you know that technology of last so many 1000 of years we are not been able to fundamentally bring about an improvement in alcohol fermentation technology that is what it says that is very interesting to observe that we still do not seem to have an answer to alcoholic fermentation not yet. Now, something more interesting also has happened let me just draw attention to that see it says you can read here saccharomyces is the organism that is used all over the world. Saccharomyces is a fungi they work well under acidic pH. So, the pH of fermentation is about 4.2, 4.3 and the pH adjustment is done by alcohol. But in 1970s they figured out that you see fungi works very slowly they we should change over to bacteria this particular zymomonas it occurs in naturally in certain ferment in particularly in some of the palm juice it occurs. So, they have isolated and all that and then they have tried to commercialize a process using zymomonas as the bacteria for fermentation. And there are several laboratory I mean even pilot plants in which this has been tried out and it is excellent performance is excellent. The reason is zymomonas bacteria the rate at which it is able to do the fermentation is much superior to fungi bacteria works faster than fungi. So, that something went wrong what went wrong what went wrong was the bacteria works best at neutral pH. And when you want to do anything neutral pH which means that you have sugar solution all nutrients all ready and it is neutral pH then invasion by several competing organisms become very high. Therefore, if to be able to operate a bacterial fermentation for alcohol requires that molasses be sterilized. Well if you do it with saccharomyces at pH of 4.2 the competition from other organisms are very low. So, alcoholic fermentations do not go through sterilization if you use fungi, but if you use bacteria you have to sterilize. So, the energy cost of sterilization and cooling it to ambient temperatures and so on prohibitive. So, while the technology worked very well economics did not favor its adoption. So, it this also failed and a lot of research in 1980s which says that you know this can be done, but it is also not very satisfactory. I went to a factory I will not name a factory in 1970s they said no no we have got a great idea we are going to do this. They said they will do cell recycle what is the idea of cell recycle r x is mu times x correct higher the cell concentration higher is the reaction rate. So, they said I have got a great answer to this problem. So, they invested on what is called as centrifuge in those days and then and did this process and it failed the process did not work. So, they are wondering what the hell went wrong you know or must fundamentally wrong mu r x equal to mu x must be wrong then correct. They put more cells into the system, but the process failed how do you explain this you understand what I am saying you have a process in which you have recycle the cells. So, cell concentration is several times more than you normally operate, but the process does not quite give you the results that you expect. Now, you got 36 you got 38 now I mean it is not anything very different. So, what seem to have gone wrong is interestingly see when you run a process particularly and say batch process or generally about batch process 20, 22 hours 26 hours as the products accumulate the bacteria starts to sort of d nature the enzymes start it starts to die basically. So, you have lot of cells in the in the output, but many of them are not viable cells. So, when you do a centrifuge and recycle you are recycling cells alright, but the viability of those cells if you recycle the very very small. So, essentially because the viability is a very small it did not give you any great result on other words pure culture fermentation see alcohol is a pure culture fermentation when you recycle you have to be sure that what you recycle is all viable, but in the in the environment of fermentation of 24 26 hours many of those cells die it does not survive and that is part of the reason why it is the recycle has not been very successful. But interestingly recycles are very very common in waste treatment see in pure culture it is not very successful, but in waste treatment if you do not recycle your process will fail you understand what I am saying how do you explain that why we recycle in a in a waste treatment plant, but we do not have great benefits in recycling in a pure culture fermentation. The answer is in pure culture the the product accumulation seems to have a very very bad effect on the organism. While in in places like waste treatment it is a poly culture where there is predator-prey interaction and therefore predator ensures that the prey is always active because the predator wants to eat correct. So, this predator-prey relationship ensures that all organisms are active. So, waste treatment recycles are critical to success of a process, but in pure culture recycle is actually not very useful. See recycle see this people have tried recycle in pure culture like alcoholic fermentation not very successful. The reason it is not successful is because during the process of fermentation of 18-20 hours viability of many of this organism starts to go down. Therefore, when you try to separate this organism after the fermentation of 18-20 hours and recycle the viable cells in that recycle stream is not very large and therefore it does not benefit you. But in recycle in in waste treatment what happens is a poly culture there are hundreds of organism working and they all work to feed on the waste that is existing. So, the predator-prey relationship ensures that is very highly active environment is established. That is why recycles are crucial for success of waste treatment and recycles are not very useful in pure culture. In in waste treatment the predator-prey relationship ensures that all organisms are always active. In the waste treatment environment we will have to spend some time on ecology which we will do when we meet. We will talk about ecology shortly in the next class. So, we will explain that, but ecology is what is crucial to success of what we call as natural processes. All natural process you need ecology alright. So, this sort of completes this part. So, we will go to the most interesting part of which is natural selection. How recycle affects predator-prey relationship? Let me draw that once again. We had a bio reactor that we had we put this back correct. Now, what happens here in the bio reactor there is lot of growth and this growth is because of what there is lot of nutrients available lot of food available and therefore it grows. Therefore, you are starting with x naught which is very small. So, there is a lot of cell growth here x which you are concentrating here and this cell growth is because of the poly culture. So, that means this this environment the multiplicity of organism present they have selected what is most useful to this environment correct and that you have concentrated here. That means you have done concentration of the most useful organism for your process which you are putting back, but in alcoholic fermentation you started with the pure culture and that pure culture in this environment it was it as it is there is no ecology. Therefore, it is getting affected because of the accumulation of the new I mean toxicity or the products and that is why it starts to die. When you do this it starts to die because so much of toxicity it will die. This is clear to all of us poly culture is what nature is all about and predator prey is what ensures highly active population. This is a very beautiful example of natural selection. See Darwin talked about natural selection and we have read in biology that natural selection at what happens in the natural environment. This is an example where I illustrate to you how this natural selection actually happens. You can do this numbers to realize how beautifully it works. The example is here there is a chemostat in which this organism is growing. It says it is got a specific growth rate mu m is so much and all that. And then this chemostat is working and as it when it is working suddenly there is an infection that happens. So, what is happening that you have this chemostat and then it is coming in and going out S naught is 100 and it is going out at 80. So, organism 1 this is organism 1 x 1 is it correct organism 1 is growing organism 2 is growing organism 2 is growing. At an instant when there is steady state is established organism 1 infects the process that is what it says. It is the problem statement clear to all of you. Your chemostat is running S naught equal to 100 and it is 80 is coming out therefore, x 2 is growing. So, you are running a process in which x 2 is growing and when the x 2 growth is at steady state suddenly an infection occurs. Infection means what somewhere from where this x 1 has come in some delta value of x 1 has come in. Your question is what will happen to the process is this question clear to all of us the growth rates of both the organisms are all those data is given here mu m is given case everything is given here is this clear. So, let us say suppose we make a plot of mu is a growth rate versus and let us say this is what this is organism 2 is that right organism 2 it is growing. Now, it says organism 1 1 is 1 is 0.5 S by 50 plus S this is mu 1 correct and this 0.3 S divided by 30 plus S is mu 2 is that clear 10 is it this is 10 is it now tell me I mean what is our understanding of chemostats our understanding of chemostat is a d equal to mu. That means when steady state is established when steady state is established d was equal to mu or in other words mu 2 was equal to d when steady state was established do we agree what is mu 2 do we know mu 2 80 grams per liter is given. So, if you plot mu 2 versus S let us say this is the curve this is 80 correct and then corresponding to that what is the we can find out S or no is this clear to all of us. Now, we say it is there is an infection due to organism 1 we should plot this I plotted that I show you in a minute that curve looks something like this this is organism 1 you will plot shortly. So, this how organism 1 looks like now what will happen you please tell me by looking at this curve what will happen is the question clear to all of us we have a steady state of organism 2 what is mu equal to can you please calculate tell me how much is it 0.26. So, it is a running at 0.26 per hour is the dilution rate at which it is working and at that instant organism 1 has come in. Now, what is the specific growth rate of organism 1 corresponding to 80 it is higher than 2 is it somewhere here S or no organism 1 will grow faster than 2. So, it is here. So, what will happen in the process substrate it organism 1 will compete for the same substrate both will start eating what is our steady state d must be equal to mu is our steady state correct d must be equal to mu is our steady state. So, what will happen to the process now d will become what is d is f by v does d change our valve settings do not change correct. So, f by v has not changed that mean d has not changed if d has not changed what will happen to the process organism 1 it will grow correct what happens to 2 will reduce then what will eventually what will happen process do 2 will get washed out you can see here that S will keep coming down and then this will be the steady state value of S. In the steady state value of S organism 2 will be knocked out organism 1 would be growing and the value of S will reduce from 80 to some lower value see I plotted it here. So, you can see here it comes from 80 to about 54 some small value that clear is the phenomena clear what I am trying to bring out see natural selection happens every minute in the natural environment we are talking about biodiversity and all that today, but the importance is right here that simply by adjusting your dilution rate you have selected a different organism. For example, it is important to realize in if you go to any you know any river which is starting as you can see as the velocity of the river changes the ecology changes the whole ecology changes. Because dilution rates are so different and therefore, the organism that can sustain will also keep on changing and this is an illustration of that is that clear to all of you why is it that organism 1 this is organism this is organism 2 is knocked out and organism 1 starts to grow you can see here it is finally, in the same at steady state there is only one organism which is organism 2 is knocked out and organism 1 starts to grow in steady state there is only one organism. We are having a chemostat in which our dilution rate is fixed now organism this infection has taken place that infection has taken place therefore, eventually what would happen eventually we said D must be equal to mu because that is the representation of steady state and our this graph says there is only one point of intersection at that point of intersection D can only be equal to mu 1 at the point of intersection D can only be equal to mu 1 mu 2 has to get out organism 2 have no chance of surviving in that environment what I am saying is that I am running my process at D equal to 0.266 under that environment when this infection occurs organism 2 will have to get out is this clear is this clear to all of you can you now see how in the natural environment this happens at every minute. So, that is why you see the organism that is populating an environment really depends upon the sheer rate of the environment see if you just look at yourself diarrhea for example, what is diarrhea what is our system doing it is increasing the dilution rate to knock out the organism diarrhea is an example where it is it is a phenomena of increasing the dilution rate. So, that it knocks out that organism it is a natural response to getting rid of the organism this is what happens in the natural environment is this clear is a phenomena clear to all of you. So, you can repeat this second part is the same exercise repeated saying that if on this side for example, what I am trying to put across different organism for example, in the second in this exercise what I have said is that organism 1 is growing at 30 not 80 at 30 and suddenly this infection occurs here the selection is for the opposite organism you know you can see here when you are starting at 30 30 see here you can see here that the organism what is selected is the organism 2 1 is knocked out the important point here is to recognize that the environment determines if the substrate concentration changes a different organism is selected see the way nature is organized you see this it is this organization that we must understand properly to understand the importance of diversity natural environment is much bigger than what I am saying is much bigger. So, this phenomena explains how things happen in the natural world now the question that we want to answer is how long does it take for this steady state to get established we know that steady state one organism gets knocked out, but how long does it take can we tell how long does it take can we write the differential equations to tell how long does it take let us write the differential equation you can solve it at home let us write the differential equation. So, I will write for organism 1 first one organism 2 is not I will write for 2 is it alright yes or no why have I knocked this out there is no the infection is momentary it has infection is not coming continuously it is momentary. So, let me write for the there are two ways an infection can occur one that you continuously put in the infection let me write I will come back to you in a minute I am taking y to be the same for both this is ok or material balance this is our material balance this is ok what we have written material balance is correct this we have knocked out this terms now in problem this in 2.1 what are the initial conditions what is the initial condition for this problem it is at steady state that means steady state value is what 80 grams per liter is the value of steady state that means at this point when this happens s naught is 80 what is x 2 is x and is going out at 80 I am talking about question 2.1 2.1 is coming in at 100 going out at 80. So, what is the cell mass 100 minus 80 is 20 yield coefficient is 0.5 the 10 grams per liter is cell mass yes or no do we all agree is 10 grams per liter is 10 mass. Now, at this instant of time this is 0 time an infection occurs what is the value of x 1 let us say some delta some 0.001 grams so much has entered. So, that the density has become 0.001 grams per liter x 1. So, much of x 1 has entered is this clear it is a momentarily now can you solve this problem now initial condition is fully specified is it clear what we are saying can you solve this differential equation because initial condition is fully specified x 2 is known sorry this is not s this is s is 80 s naught is always 100 is it clear s naught is 100 s is 80 at the time of infection x 1 became 0.001 now instead of 0.001 you can put 0.1 you can vary that number when I say how long does it take it depends upon the magnitude of the infection if it is small it will take a long time if it is large the steady state the new steady state will be reached quickly is this clear can you solve this problem and find out how long it takes for the new steady state to be achieved s or no. So, similarly for q 2.2 also you can do the same thing what is the initial condition for 2.2 our question initial condition for 2.2 is s naught is this is 2.2 s naught is 100 s is 30 correct and what is x 1 x 1 is 100 minus of 30 70 it is 35 grams per liter and x 2 is let us say 0.001 gram per liter this is the infection. So, this is t equal to 0 when the infection has just occurred similarly this is t equal to 0 when the infection has just occurred 0 plus. So, you can solve forward march is it can we solve can we finish this problem now is it is it yes or no. Now, the question I want to ask you is this is the most important question. So, how do we use the above phenomenon to manage infections in a process after all infections are something that happens to us it happens to a process if it is the biological process. How do we manage these infections what is it that you and I have learnt when having known these phenomena we have learnt this phenomena now showing that if you change the dilution rate you can knock out that organism essentially what we are saying is that if there is infection you have to change the dilution rate in an appropriate direction to knock out that organism correct. So, we have learnt that now if you want knock out an organism you change your dilution rate. So, that particular organism gets knocked out in fact industry what they do they do lot of washing it is essentially doing this only correct it is increasing the dilution rate to knock out the organisms. So, if you are talking about pure culture you know fermentation if you want to knock out an organism that is infected a process you will have to change the dilution rate and that all that substrate will have to go into your waste treatment plant. You see this is the problem of biology that you have to accept huge amount of waste that has to go into a treatment plant, but that is something that we have to accept if there is an infection I will stop there.