 Let us get going with what we said last time. Just to recap, we started with enzyme kinetics and we said that enzymes are proteins. They have active sites and these active sites get expressed by an appropriate use of the cofactor. At a suitable pH, the active sites get fully expressed and they are able to bind with the substrate to give you certain reaction rates. So, Michael is maintained this model E plus as giving E s and E s going to products P plus E. It is a very celebrated model early 1900s and it gives us a relationship of these forms of which we have spoken already that the most important thing in these that in enzyme kinetics is that the complexes are assumed to be stationary. Now, these complexes of course, there are lot of measurements available in the literature about how good are these assumptions that these complexes seems to be reasonably stationary in biology. But in commercial systems like you and I design these could be violated or in other words that the constancy of E s may not be as good as you would like it to be based on these that the form of the rate function that we get is V m s divided by K m plus s where V m is the Michaelis maximum velocity and K m is the Michaelis parameter. Notice here coming back notice here that this K 2 K 2 is the rate at which the complex gives you the product and the rate function V m depends on K 2. So, typically what does happen is that we do an experiment using certain value of E naught. But in another exercise we find that the E naught is different because the actual application requires you to use a different quantity of enzyme. See here is an example of enzyme urease and the V m is given as 1.33 and at a given at some level and you are asked to estimate what happens if the quantity of the enzyme is different you understand. This is an exercise which tells you that the experiment has been conducted at 5 grams per liter for which V m is given as 1.33. But you are asked to calculate at a different level of the enzyme is this clear this is the exercise is fairly straight forward if you recognize that K 2 is actually the parameter of the process and E naught is what you can change from experiment to experiment. So, once you change E naught V m changes and therefore, the process the rate at which the process takes place changes. So, with this quickly let us see what happens to this problem it is very elementary what do we have we have a batch equipment experiment 1 is done at E naught equal to 5 grams per liter and experiment 2 is at 0.001 grams per liter that is that is only difference and here V m is given as 1.33 mole per second per liter and what is the V m here is the question what is the V m here when E naught is 5 grams per liter V m is 1.33. So, when E naught is 0.001 what is the V m directly. So, V m V m here is 1.33 divided by 5 and multiplied by 0.001. So, V m for our case I will call it V m dash just to distinguish is 0.2 1033 10 minus 3 mole per liter per second. Now, K m does not change K m is given K m is 0.26 mole per liter. So, what is the time required for a given level of reaction in in a vessel etcetera. So, what we have to do is elementary, but only tells you the way to it is d s times d t with a minus sign which is r s that is equal to V m dash s by K m plus s. So, all numbers are known therefore, we have d s divided by multiplied by K m plus s equal to minus V m dash d t. So, this we have to integrate to give you whatever we want. Please integrate and finish it off. So, let me let me K m l n s plus s s naught plus s equal to minus V m dash V m dash t is it all right. So, we can integrate this finish it off let me let me complete this let me complete this. We get this I am just taking it forward from here only. So, it is K m l n s by s naught plus s minus s minus s minus s minus of s naught equal to minus V m t or K m l n s naught by s plus s naught minus of s equal to V m V m dash t. So, we can find out the time that is required for a given level of reaction. Now, the the reason why I have done this is to draw attention to an exercise I had given you I want to spend a little bit of time on that. See the exercise that was given to you little while ago was something like this. You had a P F R you have a huge tank here is a pump which does this. Now, this is the P F R which holds the enzyme and now here you put ammonia addition. So, that the P H is at 7.8. Now, what happens in a process is that because of this reaction the P N G the particular P N G plus water giving you 6 A P A plus phenyl acetic acid and because of this you find that the P H goes down and you have to adjust the P H. Now, if you allow this reaction to take place to a complete extent in this equipment the P H would have been very low and if the P H becomes very low the enzyme gets completely denatured. The reason that is why people take the P N G in a large equipment and then pass this through this at a very low velocity. So, that the shear rates are not very large. So, enzymes are very fragile molecules as you all know. So, if you put it a very high shear rates it denatures you have lost the enzyme cause the high. So, it is done an appropriately low shear rates and therefore, the residence times are so adjusted that the amount of reaction is also very small per pass the conversions are very small. So, the P H changes are also very small which you adjust. So, you continue this for a long time typically if P N G conversion 6 A P A process in many industries will take 8 to 10 hours per batch. So, it is done very slowly P N G is a fermentation product it comes from antibiotic factories. You hydrolyze this to make what is called as 6 amino penicillinic acid. This material is one of the very important pharmaceutical intermediate it goes for variety of semi synthetic penicillins. Ampicillin is a good example ampicillin is made using 6 A P A raw material. Now, in this hydrolysis you get phenyl acetic acid as a product as a result of phenyl acetic acid the P H becomes lower and lower as the reaction proceeds. And therefore, in the reacting environment the P H is no longer 7.8 that is appropriately desirable for the process. How do you ensure that this P H is maintained? One way is that you do not have this you put the enzyme in a basket I will draw it here this also there in the industry. So, in the industry you also see this you have a batch equipment and very very carefully designed stirrer you know this stirrer you know is and then P H adjustment. Reactor is a batch reactor where you put ammonia add to adjust the P H, but here what you have to do is that you have to keep the agitation and then this enzyme might be sitting somewhere here and it is been done very carefully. Most experience of process is that it is not easy to do this and therefore, people have thought of doing like this. So, here the shear rates are lower than the shear rates that you get here that is why this is preferred it is not that even this is available you will see it in the industry also. So, this whole thing takes about 6 to 8 hours sometimes 10 hours and then at the end of which you have 6 A P A and phenyl acetic acid which you send it for recovery. So, this is the first problem that I had set for your take home therefore, you have to see this residence time if it is tau P and this residence time if it is tau M this is very large compared to this. Therefore, this C A if I call this as C A 1 and C A call this as C A 2 you can write the C A 2 equal to C A 1 times exponential minus of K tau P this may not be a bad assumption. Then you write a material balance for this you can directly get the answers. So, the learning aspects of this of this exercise is that in a commercial process where enzymes are very sensitive this is a preferred configuration and how long you should run this reactor will come from solution of this along with the material balance solution for this which you know how to write. You see mentioned about pH effects last time let us just draw attention to this pH effects what I was saying is that enzyme ionizes and it ionizes further. And therefore, you have an E minus which is the active form of the enzyme and E minus 2 which is not an active form of the enzyme and of course, an ionized which is also not an active form of the enzyme. So, based on the equilibria for E minus and E minus 2 we have set up the final form is last time this is the final form which says that depending on the pH on the pH you will find that this ratio is affected. Therefore, the maximum velocity that you will observe will depend upon what is the extent to which the ionization gives you the value of y. So, this maximum velocity is affected by pH. So, choice of pH ensures that you are operating at the highest velocity. And if you look at some of the enzymes is available in the market this is very very sharp I mean in the sense it is almost a switch it is almost a switch in the sense that if it is supposed to be at 7.8 if you are operating at 7.9 or 7.7 you would lose considerable amount of activity. This is like a switch it is sort of it is sort of collapses, but becomes very active at that particular enzyme. So, it is important to have a very good pH regulation you will find in all these processes good pH control is a part of the process requirements. In way we said all these things we also said that the best choice of pH is k 1 and k 2 once you know k 1 and k 2 we know what is the best way to run the process. With this let us just go to this problem. So, this is an interesting problem where substrate inhibition actually affects the process. So, we want to quickly understand how do we get the form of the kinetic expression for substrate inhibition. So, this is some data I have taken from Fogler's book 1986 and this is some kinetic data S is given R S is given and the values of S for different concentrations and reaction rates are different concentrations are all given. Notice that the reaction rate goes up very high value and then start to come down. So, you can somewhere here the reaction rate takes the highest value and then starts to decrease afterwards 0.017 and it starts to decrease. So, what the substrate inhibition model see we said this and I was doing it in a hiring perhaps last time. So, I have done it again. So, you have R S I mean this is the complex R S and R S C S. So, the model as is mentioned here that enzyme complex is substrate and then the E S complex I mean complex is substrate to give you this inactive S C S complex. This how the active sites are blocked and this is the substrate inhibition model which is very popular you will find this in many places there are many examples I have taken today also. So, R S equal to 0 R S C S equal to 0 then from this equation you can eliminate and find S C S in terms of E S and S. So, it is very easy to find out what is the rate function I have done this and it looks and we do all these manipulations and all that it finally, looks something like this. So, the rate of formation of product has V m what is V m is and then k 2 and k 1 are the 2 constants and they are all given here V m is k 3 E naught notice here that as you change E naught the V m will change k 2 and k 1 are the ionization constants k minus 2 by k. So, all these constants and they come from your measurements on reaction rates. Now, what is given here is R P is given in this problem R P is given and you are asked to calculate what is the first part of the exercise is check whether it is consistent with substrate inhibition kinetic model that is first part of the exercise. The second part of the exercise is that that it is actually you do get multiple steady states in these cases. So, you will have to see what how many steady states you get and of course, you are supposed to check whether they are stable or not and so on. So, you know how to do all that. So, let us first see how best we can see whether it fits the multiple sorry fits the substrate inhibition model. How do we do this to do this what you do is the following you have suppose you do this. That means, you do an inverse transformation you get 1 by R P have done it right 1 by R P equal to k 2 by V m s k 2 by V m s plus 1 by V m s plus s square by k 1 V m s this fine. So, you do a linearization. So, 1 by R P becomes k 2 by V m s 1 by V m plus s square by all this. Now, notice here when s is very large of course, we do not know how large is large when the experiment will tell us at large s the first term becomes unimportant at large s first term becomes unimportant. Therefore, if you plot this data as 1 by R P versus s and then if you get something like this which you can extrapolate to 0. Then slope and intercept will give you the parameters correct you understand if what we are saying is right if you make 1 by R P versus s and then if you get a behavior like this. Then you can extrapolate this line and then this intercept this intercept what is this intercept value 1 by V m or in other words if our data is consistent with this kind of description only we can say this and what is the slope this is the slope V m by k 1 1 by V m k 1. Now, what is this point point of maximum reaction rate therefore, s optimum is root of k 1 k 2. So, this this also you know from the data correct you understand what I am saying. So, you have you have 3 parameters and you have 3 numbers from where you can calculate the 3 parameters is that clear can you quickly plot and then see how it looks like please all of you I plotted this see you see here see from 0 it goes to look at the data it is 0 R S is 0 goes to 0 1 7 point 1 7 it goes up and then comes down. So, the maxima is somewhere very close to see actually there were some points I could not fit it in this data there are some more points between 0 and 4 in Fogler's book I have not taken it this is clear. Now, you plot this quickly please hurry up it is not take too much time it is a very very simple problem we have to do it quickly 1 by R p. So, it is minus actually it is minus R S it is not R S minus R S see what I have written here is plus R S that is not correct minus of R S is that clear it is rate of formation of product see what is being said is that the rate at which substrate consumes the rate of formation of product under the quasi steady state approximation that we already proved. So, R p equal to minus of R S under the assumption that R S is 0 that we already proved is that is that clear it is a good question see under the quasi steady state approximation R p equal to minus of R S this is something that we have proved already. So, 1 by R p can write down all the numbers 0.17 see even if you plot 2 3 points I would suggest that you know you plot because if you take points around this region and then extrapolate also you will get reasonably good answers you do not have to plot everything plots 4 or 5 points that is good enough. So, for example, see 0 4 0 5 is 20. So, this is 1 point here and inverse 0 4 is 25. You have 2 points there actually you can just join them and extrapolate that will be pretty good is that clear. So, take few points here. So, 0 4 see this 0 5 is 1 by 2 1 by 0 5 is 20 is 1 by 0 4 is 25 1 by 0 3 5 is about 33 32 whatever. So, those points if you join directly to 0 that will give you the intercept can you read out the numbers please you plot 1 by R p versus S remember that R p equal to minus of R S under the quasi steady state approximation that is minus of R S S that minus sign is missing any numbers I get V m as 0.25 intercept is I have taken intercept as 4 is 4 is what I have taken this line when I have taken 4 is looking all right any answers what is it takes so long V m intercept is how much I have taken it as 4 ok no what is V m inverse of that is V m fine I have got 0.25 very good all right what is k 1 slope is 0.5 0.45 all right all right all right we we accept all that. So, what is k 1 I have got 8. So, V m is 0.25 and k 1 is 8 is what I have got some of you got slightly different numbers that does not matter now 0.25 is what I have taken and now the S optimum is 4 root of k 1 k 2 is 4 correct root of k 1 k 2 is 4 you can see here the highest reaction rate is somewhere here. So, that gives k 2 as 2 k 1 is 8 k 2 is 2 and V m is 0.25 this is what I have got. So, my numbers are V m 0.25 millimole per liter per minute these are the numbers what I have got slope I have got as 0.5 fine fine that is ok these are not. So, all the parameters are shall you go forward. So, parameters I will read out I will get k 1 is 8 k 2 is 2 and V m is 0.25 this is what I get this is next part what we want to do is we have a C S T R all right this and this rate law applies which means R S. So, this rate law applies this is all right what we are saying. So, let us write the material balance F S naught minus F S plus R S times V is 0 or F into S naught minus of S equal to minus of R S times V this is I call this as P I call this as Q and I both are functions of substrate concentration. Already Q versus S is been plotted already this plot P versus S can you plot P versus S. So, you have to plot this function F into S naught minus of S F is given as 3.2 S naught is also given 50. So, you can plot this plot this and find out how many steady states you get C you have to plot Q versus S you already this data is given you have to plot and P versus S it is this function F into S naught minus of S you have to plot points of intersection are the steady states the equality points of intersection are the steady states. So, please plot and tell me hurry up is this clear what we are saying S or no what we are saying is that this is material balance is this is material balance F of S naught F of S R S V equal to 0 S or no this is this data is given already I call it as Q that Q versus S is already plotted you know you have to plot and find, but P versus S is what is this function has to be plotted. So, the equality is the points of intersection wherever they are equal is the steady states. So, you find out plot and tell me how many steady states please plot and tell me how many steady states plot Q versus S the data is actually given Q is actually R S minus R S sorry. So, you plot that and then you plot this function F of S naught minus of S it is a straight line F of S naught minus of S is a straight line F is 3.2 and V is 1000 V is given V is 1000 you have to plot Q which is minus of R S V versus S that means you have to multiply R S by 1000 and then plot is that clear that is the right hand side the left hand side is F is 3.2 S naught is 50 minus of S. So, it is a slope it is it is got a negative slope take 2 points and join that is the best just take 2 points and join same graph you want to see points of intersection I see you are not able to plot is it. Say on the same can you not plot what is the problem choose a different scale what is it matter I say choose a different scale it is minus R S I told you now negative sign is missing choose a different scale on the same plot please or other alternative is make a table and you can see from the table itself that might be even faster just make a table. Now, you can read out the numbers I will write here. So, I will make 2 tables minus R S V and then F times S naught minus of S and this is S. So, just read out a few numbers they see I will just write this number itself I will just write this number itself it is 170 140 110 this is at 4812 some later values let us take some 4850 and all that we do not want too many in a 48 and 32 please tell me 32 and 48 and 36 tell me the numbers 32 is 50 32 is 50 or 55 55 and 36 is 50 and 48 is 37 is it all right F of S naught minus of S 3.250. So, at 4 46 multiplied by 3.2 is how much just 2 points will do 147 and 3.2 multiplied by 14 3.2 multiplied by 14 3.2 multiplied by 14 please plot now at least you write down plotting may be difficult write down all the numbers fill in all these numbers please what should I write here 134 and then 1 is it correct how much 57 now can we say where the answers are somewhere here I think 1 answer is here we will get 3 answers 3 steady states will come as plotting is probably the best 3 steady states will come what are the answers here how much 41 41. So, there is there is some answer somewhere here there is some answer somewhere here also 50 to 37 44 to 41. So, it will cross. So, similarly there is an answer here also somewhere is that clear see when you plot you will get 3 answers since you do not do not have graph sheets today you are not able to plot it properly. So, it is something that you can plot. So, what is the highest conversion we will get highest conversion comes from where corresponding the lowest value of s somewhere here 38 is it anyway. So, there is one answer here is it clear at all of you when you plot it you will get 3 steady states 3 points of intersection when you plotting q versus s and then p versus s and they will intersect is it clear they will intersect and you will get 3 steady states. So, you can do it at home, but this the principle is what we have illustrated here and you can do this at home. So, what we are trying to get across here is that substrate inhibition is quite common in biology and therefore, if you are running a chemostat by chemostat we mean CSTR you are likely to have problems of multiple steady states. When you have multiple steady states you have the issue of stability of those steady states we have done all that you can determine from your stability analysis which of the steady states are stable that you will do it at home and give it to me. All right we go forward now answers we will not bother about answers now you can do it at home. Let us go forward see what we are trying to do in this exercise they have taken an exercise 20,000 population of 20,000 some 6 milliliters of water has to be treated. These numbers are something similar to what happens in the IIT campus. So, I put some numbers. So, that we want to find out what is the size of the equipment that we require. So, this is the exercise we want to we want to look at. So, let me quickly explain to you what is the fundamentals that we have to understand. So, what is it that we have we have a chemostat what is the chemostat? Chemostat is a CSTR where feed comes in continuously feed goes out continuously and because of the of the processor that take place in this equipment this S naught which is a pollution level it comes down to S. Now, we have various kinds of specifications on what is the value of S should be. For example, if you look at waste waters of communities say Bombay city for example, this S naught in Bombay city we around 600 to 650 milligrams per liter measured as oxygen demand. What is oxygen demand? It is the milligrams of oxygen that is required to oxidize the waste material to carbon dioxide and water. So, if it is carbohydrate like glucose C 6 H 12 O 6 you put oxygen it becomes carbon dioxide and water. How much is the oxygen required? The stoichiometry will tell you. Now, what this says is that in Bombay city it is typically 600 milligrams per liter or more, but if you look at campuses like our own where the consumption of water is quite large our concentrations are much less because we consume a lot of water in our daily use compared to what happens to an average citizen in the city of Bombay. But even then this 200 is considered quite large because what we are supposed to put into the environment is a near drinking water quality. I mean this is what we would like to do, but we are not able to do for a variety of reasons and hopefully in few days to come we will have better and better technologies which people can afford. So, that we can reduce this to near drinking water quality. So, what has been specified here is that we should get it up to 10. Now, this value of 10 what the reason the number 10 is given is that if you if you go around the world and find out what is the specification for primary water quality for drinking purpose or in other words whenever the corporation of the world goes to supply of drinking water to the community they will look for a source where the water quality is quite good it must be good water quality it should be less than 10. That means the level of pollution in that water must be less than 10 only that source they will take they will treat this and make it 0 and then give it to you is this clear this is the process. So, what we are supposed to do is that this 200 that comes out of this campus let us say it must be made 10 before we let it out. What seems to happen in the city of Bombay due to variety of reasons the waste water of the city goes into the I mean into the ocean untreated in Bombay city that is the status all the water of the city is pumped into the sea untreated. So, the sea is in bad shape variety of reasons the fishes that we all consume is also not in great shape and so on various kinds of problems associated with the fact that we interfering with the natural water quality of the sea. So, how do we do this? So, the question here is what is it that what should be the size of this equipment. So, that it is able to come from 200 to 10 as far as let us say this campus is concerned if you are going to the corporation of Bombay they are looking at 600 or 700 they want to make it less than 10 there is a very tall order they are not able to do it for variety of reasons, but may be in course of time they will have the money to do it to do this what I have done here is I have written a cell balance. How much cells are coming in? How much cells are going out? How much cells are generated? How much cells are accumulating? Input output generation accumulation. So, now let us see what happens at steady state at steady state we knock out this term the right and side term we knock out. So, this is the steady state situation of the. So, f x naught minus of x plus mu times x v equal to 0. So, statement of material balance now what we would expect to see in this tank depending upon the technology that you employ in this tank if it is the aeration technology that happens around the world then because of the fact there is supplying a lot of oxygen here the cells would grow and cells would multiply enormously because you are supplying oxygen plus there is lot of food available previously what happened is that food was available, but there was no oxygen available. Therefore, the cell density was very small now you because of the oxygen available cells are growing and typically what seems to happen in biological processes is that we want this cell to be of our choice for example, if it is penicillin we want to have only pure culture of penicillin or in other words in biology we are interested in growing the cell of our interest. So, this feed stream is typically sterilized. So, that x naught is 0 this is what typically happens in a process now. Therefore, sterile feed is a common need in biological process therefore, we look for this material balance for the case of sterile feed even if you look at for example, waste water where the feed in the waste water this x naught is not 0, but it is very small considering the cell density here. So, the orders of magnitude different. So, to assume that x naught is 0 even in waste water is not a bad assumption in pure culture of course, you will sterilize by thermal means or it by UV radiation or whatever. So, both cases whether it is in waste water or pure culture biological process x naught equal to 0 is typically an assumption not an assumption is a requirement of the process. So, what we have done here is that we just set f by v equal to d this is what called as dilution rate the word in biology the word dilution rate is frequently used d is called dilution rate f by v has the dimensions of inverse of time dilution rate has the units of inverse of time. So, what happens if I set x naught equal to 0 and the rate of growth is mu x. So, we end up with this algebraic equation x multiplied by mu minus of d equal to 0. So, our cell balance leads to this algebraic equation x multiplied by mu minus of d equal to 0. Now, after all we are spending money constructing you know all these equipments, because we want to grow the cell correct. So, we are interested in the non trivial solution to the problem we are not or x equal to 0 is really not of our interest we will look at it later, as for the moment our interest is to grow the cell therefore, x naught equal to 0 or non trivial solution is what we are looking for is it. So, our non trivial solution is this when x is not 0 mu equal to d is our solution this is alright what we are saying when x naught equal to 0 the solution to this algebraic equation is mu equal to d now we said that mu is described by this kind of monod's model of which we talked about last time also that mu is mu m times s by k s plus s there are various other models this is the simplest form of the microbial growth model that it is available in the literature. So, we say that this mu must be equal to mu m s by k s plus s therefore, our steady state is described by this equality this is the equality which describes our steady state or the extent to which our substrate gets consumed from s naught it becomes s which is given by d k s by mu m minus of d just solving this. So, what are we done what we have done is that we have an equipment in which s naught came in s went out and therefore, this s depends on growth parameters of the of the cells in the equipment and those growth parameters are mu m and k s therefore, the output concentration is given by d times k s by mu m minus of d on other words if you chose the dilution rate at which you will operate then you can specify the extent to which you can clean the water this is clear simply choose the dilution rate then you can determine the extent to which you can clean the water is it now what seems to happen in biological processes I mean if you go to any waste treatment plant they will tell you that listen you see my organisms have got washed out this is a frequent complaint they will make you go to the plant you say the plant is not running you ask them why it is not running a plant got washed out on other words wash out is a fairly serious problem in biological process which is operating on a continuous mode. And therefore, we have to protect the process against wash out therefore, we have to first understand what is wash out then only we can protect let us look at what is the meaning of wash out meaning of wash out is that you started your process with s naught and what came out was also s naught no treatment took place process the everything got washed out therefore, you put s naught s naught came out therefore, s naught went into the sea or into the river whatever why did it happen the answer is if you put s equal to s naught here it will tell you the dilution rate at which wash out occurs correct let us do that put s equal to s naught here. So, what is the wash out dilution rate so, what is d w equal to mu m s naught s naught k s plus s naught. So, what we are saying is that if you are given the monod parameter and if you know the concentration at which your waste water is coming you can immediately tell the dilution rate at which the wash out will occur which means what to protect your plant against wash out you should not allow the dilution rate to become as high as this. So, what is dilution what is d w d is simply f by v the question is why does this dilution rate becomes as high as d w it becomes very high because if it is waste water treatment you find that between night and day there is a great difference in the consumption of water during the day the water consumption is f during the night the water consumption may be 0.2 or 0.3 f you see therefore, there is a great difference in the amount of water that flows into the plant during the day compared to during the night therefore, this variation if you have not properly accounted for in design you will find that the dilution rate during the day will be so large that your system will get washed out. So, how do you protect your plant against wash out people say that you know you build a huge tank which is called as the you know holding tank. So, large that you know even if there is great variation this tank is able to hold enough water. So, that this does not affect your process, but if you are building a plant for the city of Bombay where we consume 3000 million liters per day you cannot build a holding tank correct it is not possible to build a holding tank you have to do something else let us see what we can do. So, what is done is the following. So, you have a bio reactor then you have a settling tank what is this tank this is a settling tank what is settling tank do you get a settle layer which is got much higher cell density and clear water where the cell density is low correct or you can put a centrifuge here you can put a centrifuge. So, that you get a very thick microbial biomass which you can recycle and then clear water you can dispose you understand this is what is done in the process industry you put a settling tank because they that cheaper and etcetera. There are some places where they use centrifuge also. So, you have a settling tank which gives you dense biomass somewhere here somewhere here you get dense biomass and part of that you put it back into the process is this clear now we want to see how the effect of this recycle how does it help us what does it do we would like we want to know. So, for that I have written a cell balance what is this balance see whatever f is coming in at s naught and x naught what goes out of the bioreactors f s x and s and after the settling tank what goes out is h x and g x and what is recycled is a times f s and g x I mean this is a way of trying to represent what our process is all about is that clear this is our process that after the bioreactor whatever is the water coming out a fraction of that you recycle and in this settling tank the settle sludge which is got dense biomass part of that you recycle is that clear. Now what I first written is a balance a times f s a times f s plus f equal to f s can I write this this balance acceptable f is volumetric flow f is meter cubed per hour. Now this volume balance is acceptable if the density change is not significant volume balances are acceptable if density changes are not very significant in biology this cell here might be of the order of about 0.2 percent this is the kind of cell density here generally. So, this is not essentially water you can say. So, there is not great difference in cell density therefore, we accept this volume balance as ok as far as waste treatment is concerned. So, there we get f s is f by 1 minus of a it is ok. Now let us see how it helps us. So, I have done this cell balance in some detail now. So, what is the cell balance please look at this bioreactor somewhere here this is the bioreactor you can see input f it has no cells I have knocked it off output is f s times sorry sorry this a f s times g x input output is f s times x plus mu x equal to 0 is it alright I will go through it once again. See this settling tank essentially separates that water into two fractions one is a concentrated cell fraction which is g x a light fraction which is h x. So, this concentrated fraction we recycle. So, how much we recycle a fraction of f s we recycle is that ok it is the process clear to all of you. So, this material balance now can see here input I have knocked it off and there is input coming from here I have written it as a f s times g x. So, in cell I am output is f s times x and then the rate of generation of cell equal to 0 is it alright. So, let us go forward now. So, if you simplify what do you get a g f x f of x. So, this simplifies like this do you all agree with this see this thing simplifies like this mu x common and then I have knocked out divided by v got dilution rate here f s I have written as f by 1 minus of a is it alright with everybody. So, cell balance gives you mu minus of 1 minus of a g by 1 minus of a multiplied by d multiplied by x. So, I have written this like this where this 1 minus of a g by 1 minus of a I have called as b is called as concentration factor in biological literature b is this term 1 minus of a g by 1 minus of a I have just denoted it as b. Now, notice here that this is all decision variables all these are decision that you and I will take. So, b is essentially in your hand you determine the value of b correct is this clear to all of us you will determine the value of b. Therefore, our material balance process or material balance now becomes this cell balance I am sorry cell balance is mu times b d multiplied by x equal to 0 previously we had only d mu minus d equal to now we have term b multiplied. Now, what is the value of b what is our estimate what is b less than 1 greater than 1 what is it b is less than 1 why is it less than 1 a is less than 1 g is greater than 1 is that clear g is greater than 1 a is less than 1 that is why 1 minus of a g by 1 minus a is less than 1 is that alright we agree. So, b is less than 1 mu minus of b d times x is 0 if we want non trivial solution because after all we are spending money to grow the cells anyway therefore, mu equal to b d is our solution therefore, d this equality gives us the value of s is given by this relationship previously we had d k s and mu m actually previously this b was not there on other words what happens is that the dilution rate that we had previously calculated that gets modified because of the recycle that is what it means now b is less than 1 therefore, the value of s is given by this relationship this is with all of us now what is d if I ask you what is d w from here if I ask you what is d w wash out dilution rate what will you tell me it is 1 by b mu m s naught by k s plus s naught s or no what is the value of b less than 1 what have we done by choosing the b appropriately we have protected our system against wash out is that clear what we are saying now we do not have to build a huge store I mean equalization tank it is not possible when the time when the flows are very large 100 200 500 milliliters per day building a storage tank is not possible, but this is possible you understand you can choose b appropriately to protect yourself from wash out less than 1 d w has increase yes which means what that you have protected it against wash out that means previously let us say between night and day the value of f change by 3 times you choose b appropriately so that d w becomes 5 times the average flow, so that it does not get washed out is that now let us see what is the effect of this on the productivity for cell let us do a cell productivity what is cell productivity can I say this is cell productivity input output so much is produced so much is coming in divided by volume what is this term I put a this plus here minus here. So, it becomes 1 minus of a g by 1 minus of a so this becomes b d x this is all right please tell me productivity for cell is b d x that means in the recycle scenario the because b is less than 1 the amount of cells that is produced per unit volume per unit time is less. So, what we done we have lost in terms of productivity, but gained in terms of wash out dilution or in other words what we compromise see whenever we have a problem we compromise we gain at some cost this is the cost and this is the gain. This is the gain and this is the cost is that clear what we are saying is it what we are saying that what we have ensured is that productivity we have sacrificed and to safeguard the system against wash out is this clear to all of us. Now, let us look at this problem now now quickly all the data is given tell me what is the size of the equipment that is required for let us say a campus like IIT. So, IIT campus see I told you know we are we generate a lot of waste water I mean consumption is also very very large we are 6 is more than 6 million now it is much much more than 6 million. So, 6 million liters 20,000 population and so on please do a calculation quickly. So, we have to look at the answer is this what is s required 10 what is b given is 0.3 k s is given everything is given s or no. So, we can calculate what is the dilution rate at which we must operate s or no tell me what is the answer what is the dilution rate at which we must operate s is 10 b is given k everything is given or you can calculate d from this expression is even easier it is easier to calculate from here you can calculate from here mu m s is given k s divided by d. So, d is simply 1 by b times. So, this is what this 10 10 20. So, 0.3 by 10 0.15 divided by 0.3. So, 0.5 0.5. So, shall we say that dilution rate is 0.5 per day. So, what is the volume flow 6 million means what 6000 cubic meters per day. So, what is the volume of the equipment v equal to f by v is d. Therefore, v equal to f divided by d f is 6000 d is 0.5. Therefore, it is 12,000 cubic meters. So, we need an equipment volume of 12,000 cubic meters is this clear. So, you will see in waste treatment the equipment size is a very large the reason why it is very large because our reaction rates are very small. Why is reaction rate very small? Why is reaction rate very small? It is very small because we want to clean it to very high level of purity and we are operating a stirred tank we are operating a stirred tank which is operating at the exit concentrations. Do you appreciate this problem? Please this is the problem that we face and this is part of the reason why the cost is so high that we are not able to afford is this clear. Because we are operating at the exit concentration of 10 milligrams per liter concentrate reaction rates are very low plus we have to operate because of wash out and all that we have to recycle which makes it even worse. And as a result for a 6 milliliters per day plant you require at 12,000 cubic meters of volume 12,000 cubic meters means typically it is 1 meter deep. So, you are talking about 1.2 hectares which is our football field. You can see football on one side in Jim Khanna there is football field on one side of the the athletic ground that is about 1 hectare. So, that is a kind of size that is required and in a Bombay city you know the cost of land is you know such that this 1 hectare might cost you some thousand crores you know. So, it is cheaper to throw it away into the sea that is the status. It is unfortunate, but we must find a way around it. Just I just want to spend one few minutes on something else let me see if I can get hold here it is. So, what it says see alcohol industry the context of alcohol industry is that government of India wants us to put up to at least 5 percent alcohol in gasoline. And for a variety of reasons global warming and so on import reducing imports and so on. So, this is just some data on alcohol fermentations that you will see around the world most of these numbers will apply to wherever you see it does not matter. It says the rate at which saccharomyces is able to consume the hexo substrate is given by this kind of expression whereas, mu m s divided by k s plus s that is the usual form of the monauts law. It is multiplied by this term which is the exponential of minus k p where k is a constant and p is the product if p refers to alcohol. So, what we are trying to say here is alcohol fermentation from hexo sugar that you will see around the world will face this kind of rate function problem which they must factor in design. So, what is the if that means product inhibits the rate at which the reaction can occur. Here is an instance of product inhibiting the rate at which the reaction can occur is this clear is it ok. Now, under this situation if you ask if I ask you what is it that we can do from the point of view of a process design. What will you say product is inhibiting the process it is not letting it go forward product accumulates in the equipment and therefore, as the reaction proceeds the rates become smaller and smaller. If you are operating a stirred tank or a chemostat in the biological literature the output concentration if you are going let us say from some concentration of 100 milligrams per 100 grams per liter to 10 grams per liter as an example CSTR operates at the exit concentration. Therefore, the concentration of sugar at the exit is only 10 that means this s is 10 correct. This s is 10 this s is 10 what is this p what is the alcohol product how much alcohol has been produced if I ask you what will you tell me. We have started at 100 and it says alcohol formation is 40 percent is per gram of substrate that means 40 percent of glucose consumed is alcohol. So, if you have gone from 100 to 10 which is 90 90 is consumed. So, 40 percent of 90 is 36. So, 36 grams per liter of alcohol is sitting in the broth and the effect of that 36 grams per liter is minus k multiplied by 36 grams per liter that is the effect which reduces the reaction rate is this clear. So, this problem is what is significantly retarding the progress of alcohol industrial productions around the world. We do not know how to handle this problem we do not seem to know how to handle this problem. There is a lot of history to this I will talk about this when we meet next time there is a very interesting history to this and is worth knowing this history because if some of you go into this research this history is useful because in at least we do not make the same mistakes that our forefathers have made I will stop there.