 So, we look at a problem in environmental reaction engineering and the background to this is as follows. Let us look at what happens in real world. See all over the world there is a general tendency to put waste water particularly sewage into the river and do your design. So, that the river is able to take care of the pollution in the water. This is a practice that sanitary engineers have evolved over a long time and there are some guidelines which you are supposed to follow and this whole oxygen sag analysis tells you what those guidelines are and what is it that you are supposed to do. The general approach here is that the river comes with relatively clean water and the river has a natural mechanism for oxygenation and therefore, that oxygen capacity that the river has you are able to utilize to treat your sewage. This is the general thinking that river has lot of water. The river water is relatively not polluted and there is natural re-aeration capability that the river has and consequently you are able to treat sewage and there are some numbers which are to take care. So, that the sewage is treated in an appropriate way. What you want to do is to do this exercise we do it for a steady state assuming that the river flows are known and the sewage flows are known and at steady state or inputs outputs generations and accumulations. This accumulation is taken as 0. We take this as 0. Now, what is this balance is for the amount of oxygen that you are able to supply to the water in the river basically that is the balance. On other words how much of pollution load is coming in, how much of pollution load will go out and then how much of oxygen you can supply, how much of oxygen will get consumed and that balance if it is at steady state goes to 0. That is the balance that we want to write and understand how to write this. So, this is what I have written. Let us spend some time on this. We are taking an elemental volume. This is the river flow at velocity u input output generation accumulation. So, I have taken an elemental volume. So, so much what is C? C here represents the oxygen that is present in the water. What is the dissolved oxygen in water? So, C is the dissolved oxygen in water. u is the velocity of water. A is the cross sectional area of the river at x and x plus delta x. The first term here, this first term here, what is this first term? It represents the amount of oxygen that is consumed by the respiratory organisms in water which uses the pollution S 1. What is this term? S 1 times K 1 times A times delta x. This is the amount of oxygen that is consumed by the respiration in the of the organisms which using this pollution S 1. For example, you know when we throw let us say carbohydrates into water, carbohydrates you know it requires certain amount of oxygen and that S represents the oxygen demand of that water. Suppose, I give you a waste water and ask you what is this oxygen demand? We said earlier also we do a simple test to find out how much oxygen is required to oxidize it and that determines the amount of oxygen demand of that water. So, S 1 is the oxygen demand of that water and K 1 is the rate constant the rate at which this gets consumed. Therefore, this term represents the rate at which oxygen would be consumed by the pollution load in water. The next term which is K 3 times C S minus of C this is the driving force that means solubility of oxygen in water is C S, oxygen concentration in water is C. So, this is the driving force this is the rate constant for re-aeration. What is K 3? K 3 is rate constant for re-aeration. So, so much of oxygen gets consumed because of the pollution load, so much of oxygen is supplied because of natural re-aeration. Is that clear? Is this term clear? This is the pollution the oxygen consumed by the pollution load in water. This is the oxygen that is available to water because of natural re-aeration. So, this now this term R p, R r p and R s. R p represents the amount of photosynthetic oxygen supply. R p represents the amount of photosynthetic oxygen supply. R r p is the respiratory oxygen consumption by the natural organisms of water. On other words in the absence of this term for example, the river would flow which means there is natural re-aeration and that natural re-aeration would be consumed by all these activities. Only there is a natural respiration photosynthesis natural respiration there are lot of sediments in water which will also consume oxygen. So, what we are trying to say is that in the absence of adding sewage into water what you are having is that there is a natural re-aeration from atmosphere. There is oxygen supply because of photosynthesis and there is oxygen consumption because of respiration in the absence of our pollution load. Even when there is no pollution load there is a natural pollution load because of various organisms present in water. And this is the sediments which are present in water is this clear the statement of material balance is what is important the rest is very straight forward. The statement of material balance is that there is pollution load in water and that we are adding this sewage is because of this sewage. So, this is the amount of oxygen that is consumed by that sewage. This is the oxygen that is supplied because of natural re-aeration what is meant by natural re-aeration this water of the river is in contact with the atmosphere. There is atmospheric turbulence and as a result there is oxygen transfer to the river that is this term. Now, there are these three terms here which is oxygen generation due to photosynthesis. This is oxygen consumption because of respiration in the natural environment. This is the oxygen consumption because of various sediments that is present in the river basin. So, even in the absence of suppose even this you are not there you have all these terms which means that there is natural re-aeration there is natural photosynthesis there is natural respiration there is natural sediment respiration. So, when you add this is an additional term the object of this analysis is that what is the effect of this additional term on the water quality of the river is this clear this is the problem we are trying to solve. Now, these numbers k 1 k 3 etcetera are fairly well documented. For example, different climatic regions of the world there are lot of numbers which are available readily. This term r p r r p and r s lot of material is available what is important to recognize is that this term is generally considered as 0. On other words what we are saying is that the oxygen that is supplied by photosynthetic organisms of water is essentially to serve the respiratory needs of the water environment is that clear the oxygen that is produced by the photosynthetic organisms of water is the oxygen that is used up by the respiration of the natural organisms of water indicating that really speaking the natural environment really does not have much surplus oxygen natural environment does not really surplus oxygen. Because this r p is essentially meant for the respiration respiratory oxygen demands of r r p and r s, but because urbanization and so many other issues what we have said what sanitary engineers have said is that we can put certain amount of waste into our water bodies and that would not harm the water bodies. This is the argument that our friends in sanitary engineering are saying will see what kind of results we get. Now, we can represent equation one whatever that in this form can knock off the appropriate terms a goes off a goes off. So, is this what I have done I had knocked out a and all that I put it in this form does it look all right. So, this is the statement of material balance where c represents the oxygen concentration in water k 3 the re aeration constant k 1 is the what they call as I mean b o d consumption rate constant and beta is this term beta is this term is that clear is you cannot read this this whole term is beta is written as minus beta it is minus beta. So, as you can see here if you divide x by u it becomes residence time and the reverse can be considered to be in reasonably in plug flow the dispersion coefficients are not very large. So, it is a first order differential equation d c d tau that is what I have got here. So, the equation that describes what goes on in the river is this make sure all the numbers are minus minus minus plus seems all right. So, our differential equation is the rate of change of dissolved oxygen in water with respect to distance is given by this equation. Now, I put the sanitary in literature uses this symbol to call it as oxygen deficit this is called as oxygen deficit psi is a measure of the deficit means what is the difference between saturation solubility and the present concentration put it in that form our equation looks like this. So, this is what we have to solve what is our what is tau equal to 0 not tau equal to 0 tau equal to 0 do we know this what is the initial state as the river starts off do we know what happens this is where the waste is entering the river do we know what happens here. Generally, you should be able to take a sample and find out what is the oxygen concentration c i and what is the pollution concentration s 1 both are something that can be you know determined experimentally. So, the initial state of the system is fully known c i is known s 1 is known because all the numbers are known at the x. So, what is being said here is that the initial conditions psi is fully specified therefore, it can be integrated to find out the solution can you please quickly solve and see whether the result that I get is satisfactory please quickly solve this quickly. So, this is the solution that I get it does not mean that it is right. So, please confirm to me it is ok. So, this is the solution I get please see if it is ok this is my solution please see if it is ok. Now, I want to draw a tension to this term please this is important term see this term k 1 s 1 this term s 1 is written as s 1 0 e to the power of minus k 1 times tau. Why it is written like that is that the assumption is that this material which is the pollution load in the water it will decay as per the rate constant k 1 in the time of residence tau. On other words during the period of tau of residence time in the river this would consume. So, much of oxygen that is what it decays like this and therefore, it has an effect on the oxygen. So, what we are saying is that this term this s 1 which is the pollution load in water it will decay as per the exponential rate law which means that s 1 would change as s 1 0 exponential of minus of k 1 t where k 1 is the decay rate constant for the pollution load where s 1 is the pollution load. So, this comes from our first order law that if s 1 0 times e to the minus k 1 k 1 is a rate constant for the decay of pollution. So, on other words what we are trying to say here is that the oxygen that is present in water it will decay and that is because this is consuming that oxygen. And this is consuming the oxygen at this rate where s 1 is going to decay like this s 1 0 e to the minus k tau and beta is generally taken as a constant because it does not change if the environment is fixed beta is fixed. So, our solution looks like this my integrated form of the solution please tell me where it is. So, k s 1 is replaced by s 1 0 e to the minus k 1 tau beta is a constant it is taken as constant. So, I put the integrating factor and this is the solution is it what I have done my friend is that I have put x by u as tau is it x by u I should have mentioned here. Let me write it here tau equal to x divided by u that is why I have put it in that form in the next page. This is now can we go forward integrating factor I put it in this form I integrating both sides shall we go forward it is now you divide throughout by e to the minus k tau and you get the final form. So, the final form I get is this tau equal to 0 sorry tau equal to 0 psi is psi i and how do you find out the value the constant of integration how do you find the constant of integration tau equal to 0 psi is psi i. What is psi i you ask yourself psi is defined as c s minus of c this is known because you take a sample you can find out the oxygen level in the water at the point where you are starting your tau equal to 0. So, c is c i is known c a saturation solubility is known therefore, phi i is known is that clear on other word initial deficit in that water is known because c s minus is called as deficit w psi i is known from our initial state. So, psi i is known therefore, the constant of integration is given by this and solution looks like this. So, if you want to put our waste into our reverse we must look carefully at this equation. So, this tells us what will happen to our river when we put waste into our reverse that clear. So, what are we saying that psi which is the oxygen deficit psi is oxygen deficit c s minus of c psi is c s minus of c that means saturation solubility minus the actual concentration that is psi that is given by k 1 s 1 0 is the what is s 1 0 s 1 0 is the concentration of pollution after it is gotten mixed with the river our model is this pollution is coming in here here it mix up mixes up and the concentration s 1 0 represents the combination of river water as well as the waste water. So, this is s 1 0 is that clear. So, s 1 0 is known because the river flow is known amount of waste water coming is known therefore, from a material balance you can calculate what is s 1 0 is that clear. Now, k 1 and k 3 are k 1 is the constant that is responsible for the d k of the pollution k 3 is the re aeration rate constant which comes from our experiments that is available for that environment tau is the residence time of the of water in the river. So, all this equation everything is known therefore, if you put in all the numbers you should be able to tell what is the deficit of oxygen in our rivers they are clear. Now, the one point that we must bear in mind is that this saturation solubility of oxygen in water is a very strong function of the environment. For example, in our kind of climate where the temperatures are 30 plus typically solubility of oxygen is about 6.2 or 6.3 milligrams per liter well if you go to coal countries where the ambient temperatures are typically 14 and 15 the solubility is a 10 milligrams per liter. So, 6 and 10 almost it is not double, but significantly higher. So, you will find that the aquatic organism that live in very rich oxygen and very lean oxygen they are not the same. So, you should recognize their fact that you see in the natural environment is a great biodiversity depending upon the environment itself. So, if you take a if you take a fish which is suitable for a warm climate and put it into another river which is it does not work out very well because it is not suitable suited to that kind of environment. So, we have to appreciate all those features when we try to design systems like this where we put our waste into our rivers. Let us ask a small question I just reanalyze this in this form. So, that we understand each term a little better. So, this is how the solution is available in the literature The first term the same thing is arranged slightly differently. What is this term? This is due to pollution. While the second term is reaeration. So, this whole term essentially tells you what is the effect of the pollution that you have put in. That means what is the oxygen demand that your pollution load is putting into the water. This is the first term. The second term which is beta divided by k 3 and all this. This is the term which comes from the fact that the natural environment there is lot of load due to the natural environment itself respiration sediment and photosynthesis. All this we taken together is what affects this term. The third term is what is called the initial deficit. That means when we mix the waste with our river you start with some initial deficit. It is not saturated water it will be some deficit. So, essentially what you will see is the net result of what we have put in terms of what is the natural environment and what we start with the initial deficit. There are three components to what will affect your deficit for a long time for a long time. I mean this goes on even today situation the western world is slightly different where the densities are not very large. But the densities in this country is very very large. And therefore, to assume that our rivers have the capacity to receive pollution is generally not correct. Number one there is not enough flow in the river. So, our river flows second rivers flow only for about two months of the year. Because the rainfall is only for 100 days not even 100 days may be 50 days in a year. So, the second rivers do not have much flow. The Himalayan rivers have flow, but flow varies a lot. And huge amount of water is taken for irrigation other purposes the river flows are very low. So, what is being said in most places nowadays is that rivers do not have adequate oxygen to manage your load. That is the kind of argument people are giving. Therefore, we should not put our waste into our rivers that is the kind of arguments. But when you do some of these calculations I sorry exercise here. So, if you want to look at the first problem. So, that we can appreciate what I am saying. This is the river 30 million gallons per day 30 million gallons is what 1 gallon is 3.8 liters. So, 100 million liters 100 m l d Bombay city is about 4000 m l d. So, you can see 100 m l d is not very large, but it is not very small either. So, this would be a small town 30 m g d will be a small town cities like Pune is about 6 700 m l d Pune. So, it is not very small, but substantial. This sewage goes into a river with the flow of 100 10 cubic meters per second. Now, 10 cubic meters per second is not a small flow. You will find most decant rivers most decant rivers. I mean when I say decant rivers I am not talking about Godavari and Krishna they are very big rivers. I am talking about small many of the small tributaries you know rivulets during the monsoon they will have this kind of flow. The river will have much bigger flow, but many of the small say Bima for example. This will have this kind of flow during the monsoon for 50 days or so. Then if you are going to put your waste into this at that time these calculations would be relevant. But if you are going to do it at any other time these calculations will tell you that there is not enough flow and these calculations are not very relevant. So, we are looking at an instance where there is lot of flow in the river number 1 to see what happens when you put sewage into the river at a time when there is plenty of flow. When there is very little flow it is a different scenario which we will do later. So, what it says is that stream velocity is a given 5 days sewage what is meant by 5 days sewage BOD means I am sure you understand this. See when you try to determine pollution load in water we said there are 2 ways by which pollution load in water is determined. You take that water and try to oxidize the dissolved organics using potassium dichromate. It is a standard oxidation. Now amount of potassium dichromate that is used for oxidation of the pollution is expressed in terms of milligrams per liter. How many milligrams of oxygen is required per liter of the solution for oxidizing? This is fairly standard, but BOD actually is a slightly different technique. In BOD what you do is that you take let us say a liter of water which is saturated with oxygen. You inject your sewage into this water. So, you know initially it contains so many milligrams per liter of oxygen. You wait for 5 days and you measure how much of that oxygen that you have put in got consumed because of the sewage that you have put in. This is what is called as BOD biochemical oxygen demand. That means oxygen that we have put in has been consumed by the sewage organisms and that is what determined by BOD. And this is done after a 5 day period in the sense that you can wait for 10 days, 5 days, 3 days, 2 days that choice is yours depending upon how long you can afford to wait. The standard technique around the world is you should do a 5 day BOD. That means you take the water saturated with oxygen, inject your sewage, wait for 5 days and measure the difference in concentration of oxygen. Whatever is the oxygen that is consumed divided by the volume of water will be the amount of oxygen consumed per liter of that water. That is how they express BOD. Now, when you say BOD of 200 milligrams per liter which means that it has 200 milligrams per liter of oxygen demand that is coming from sewage organisms. It is coming from sewage organisms. Is that clear? Let us just try to make a quick calculation. Most of us excrete typically about 80 grams per day. Most of us. We are supposed to consume something like as per WHO norms 180 liters of water per day. I mean consume means in various ways drinking, bathing, washing, etc. So, 80 divided by 180 is the likely pollution load in that water. How much is that? 80 divided by 180. Something like how much is that? 80 divided by 180. So, 400 milligrams per liter is the kind of oxygen load that we should expect in the waters that you and I dispose. But if you consume instead of 180, 400 liters as we do in IIT Bombay for example, that is our average highest in the world. So, that becomes 200. What I am trying to say is that this figure of 200 milligrams per liter depends upon amount of water you consume. There are places in India where the consumption of water per day is only 50 liters. So, therefore, the load on that pollution would be in 50 liters 80 grams would come. So, that will be 1600 milligrams per liter. What I am trying to say is that if you go to the villages of this country, where the availability of water is low, the pollution load is very high. That means, so much of oxygen is required to oxidize the wastewater. So, you cannot look at wastewater in terms of MGD only. Volume alone is not enough. How much pollution load is present in that water is equally important because that much oxygen will have to be provided for it to be oxidized. So, when people say 30 MGD containing 200 milligrams per liter of pollution load. These are important figures that we should bear in mind. Now, tell me if the river is coming at 110 cubic meters per second, it is mixing with 30 MGD, what will be the concentration of pollution at the point of mixing? At the point of mixing, what will be the concentration? Is that question cleared all of us? Let me repeat the question. The question is the following. Question is that the wastewater and the river is mixing here. So, what will be the concentration? Wastewater is coming here. This is the river. This is the mixing here. What will be the value of S? Pollution load here. How will you calculate? To calculate this, you can see here 30 MGD, 30 1 gallon is 3.8 liters. So, you know the total volume flow. It is coming at 200 milligrams per liter. So, you know 30 multiplied by all those. Then this particular river water is coming with the BOD of 1 milligram per liter. Is that clear what I am saying? Can you calculate what is the value of S? I have done this calculation here. See, I have done this. I hope it is correct. You can tell me whether it is correct. I have calculated the DO in the mixture and BOD of the mixture. Is this correct? Please tell me. What is the dissolved oxygen in the mixture? It says solubility of oxygen is 9.9. Now, please tell me. What is the DO in the mixture? What is the BOD 5 of the mixture? This is the calculation I have done. You please tell me whether my calculation is right. I have done with 9 because 90 percent of this, what it says here, 90 percent saturated. See, the river does not come 100 percent saturated. It comes with 90 percent saturated. So, I have taken 90 percent of 9.9 as 9. Is this clear? What we are saying? We want to calculate what is the DO in the mixture. We want to calculate this. How do we calculate this? How much of oxygen is coming in divided by the total flow? Is that okay? Is this number correct? 7.95. Can you please calculate and tell me? 10 cubic meters per second. This is cubic meters per second. This is okay. There is only 1 S, 2 S. Anymore? Is it alright? So what are we saying that? See, most of this data refers to coal country. I have taken this data from coal countries. So, why this solubility of oxygen is very high and then therefore, your DO in the mixture is 7.95. Now, I want to calculate what is BOD 5 of the mixture. That means, how much this pollution is coming in here? There is some pollution in the river itself. So, when the 2 mix, what is the BOD 5? I have just calculated here. Tell me whether this is okay? I get it as 27.7. 30 mgd, 3800, 200 milligrams per liter and this is the river total flow denominator. See, most of this calculation I have done mentally. So, I am not too sure whether this is right. 24. Anybody else? Is there consensus? Everybody is agreeing with this? Is it clear what I have done? See, this total amount of pollution coming in, this is the pollution in the river divided by total flow. This is okay. We will go forward. DO is dissolved oxygen in the water. See, aquatic life requires oxygen. Fish, for example, requires dissolved oxygen. So, this tells you how much of oxygen is present in the water. This tells you how much pollution is in the water. Is that clear? Let us ask a few questions now. What do we expect? What do we expect to happen to this river as it goes along? What will happen to psi and what will happen to S? What is our understanding? What will happen to psi? Psi refers to oxygen deficit. That means, how much below saturation it is there? What would happen to psi? It will go up or come down. What do we expect? Now, we should expect that because of pollution, the oxygen concentration will come down. Because of re-aeration, the oxygen concentration should go up. The net result would be that for some time, the concentration will keep on going down and then it will come up. In other words, the sag will keep on sagging and then it will come up. This is what we would expect because we have put a lot of pollution load. As a result, initially the pollution load will consume the oxygen and therefore, it will keep going down and then it will come up. So, the concern of people who are looking after rivers and lakes, etcetera, is that the extent to which it goes down should not be detrimental to aquatic life. See, if we allow it to go down, yes, how much do we allow it to go down? Now, if there is no oxygen, of course, fish will die. Whether the fishes can live with 1 milligram per liter or 2 milligram per liter or 3, we do not know. But our friends in environment would know. They will tell, we should not allow it to go beyond whatever is the critical number that is specified for different environments. The object of this analysis is to determine what is the likely sag that will be experienced if you put this sewage into the river. Before you do it, when you do your calculations, you will know. Then accordingly, you can do your design, making sure that the bad effects will not follow. We have been talking about oxygen sag analysis. We have derived the equations that describe the consumption of oxygen in the river and so on. Now, just to summarize, we just put it in the context saying that if sag is the oxygen sag and this equation represents what happens when there is BOD alone acting on the water. That means d psi d tau, but k 3 tau equal to k 1 s 1 plus beta when there is BOD alone. What are these terms? Psi is the oxygen sag, tau is the residence time in the river, k 3 is the reiration rate constant, k 1 is the BOD removal rate constant and beta is the oxygen load that is coming from respiration and sediments. This consumes oxygen and photosynthesis, supplies oxygen and generally this difference is denoted as beta. Now, you can have a situation in which you have only COD that is entering the river. Now, as you all know, most of the industrial wastewater, the BOD is very low. Most of it is COD only. Therefore, there could be situation entering the river is only COD which is s 2 or it can be only BOD which is s 1 or in the mixture of the two both are possible. We will consider all these situations as we go along. We have already solved this case for the case of BOD alone. We gave the solution already. I will not do it again because we have done it. But what is important to recognize here is that the oxygen sag is actually affected by three important factors as you can see here. One is the effect of BOD that means what is the exertion due to BOD to bring about a reduction in to sort of change the oxygen sag. This is the effect due to photosynthesis, respiration and so on. This is the initial effect. That means at the point of entry because of the waste coming in, there is an initial deficit. On other words, what we are saying is the effect of BOD on the oxygen levels in the river, this is the oxygen sag. The oxygen sag is affected by BOD exertion, affected by initial deficit. It is affected by the net load due to respiration, sediments and photosynthesis. Now, if instead of BOD, we had COD, this is the same in the sense that all the effects are due to COD. This is due to COD exertion. This is net load due to natural biology and this is the initial deficit. So, the form of the two equations are identical. Please notice wherever S 1 0, it becomes S 2 0. There is no difference. The form is the same. Only thing is that here it is k 1 rate constant. Here it is rate constant is k 2. So, the form of the expressions are identical whether it is BOD or COD. Only when they are mixed, the expressions are slightly different. We will come back to that in a minute. Now, we have shown this. I will not show this again. We have already shown that for the case of whether it is only BOD or only COD, that the point at which the maximum sag occurs is given by this expression. I have not derived this. I have left it as an exercise to the class. I request all of you that you please substitute and convince yourself that this is correct. If necessary, I will take it up in a tutorial class. So, the point of maximum sag is given by this and the value of the maximum sag is given by this. That is what is the important point. So, what have we said so far? What we have said so far is that we have set up the equations for the level of oxygen in the river after there is some amount of contamination due to pollution. Then, we expressed this effect on the oxygen in the form of oxygen sag, where sag itself is defined as psi equal to Cs minus of C, where C is the saturation solubility of oxygen and C is the local composition. Then, we said that the oxygen sag in the river at any position given in terms of residence time can be related to the system parameters, which we have done. Now, we are now in a position to actually see how best we can use our results to understand how things perform in a river. So, we want to see how our formulation is able to explain what goes on in a river. Now, the question that may be relevant to all of us is how is oxygen sag analysis of river relevant in chemical reaction engineering? It is something that we would all like to know. It seems to be a problem in environmental engineering. What I am trying to put across here is that whatever happens in a river is due to a chemical reaction. What is that chemical reaction? There is a chemical reaction, which is consumption of oxygen of the river due to the biological reaction, which leads to growth of that organism. Now, that means the consumption of oxygen by the biology is a chemical reaction. Now, that oxygen that is being consumed must be replenished. That replenishment comes by a physical mass transfer. On other words, here is an instance of a reaction problem, where there is mass transfer and chemical reaction. It is a very interesting example and how mass transfer and chemical reaction interplay to ensure that the river has a certain oxygen level that we see. I mean, our interest in looking at this kind of reaction engineering problem is to see how best can we look after our natural systems. How best we can manage these natural systems in the context of so much of pollution load coming from human activities? How best can we rejuvenate these systems, so that they remain in good shape, so that they serve our livelihood purposes as well as we would like it to be. In fact, it is the most important point that we want all our natural systems to be in good health, so that they serve the livelihood purposes not only for us, but for all the living organisms around us. So, our object of looking at problems like this is to apply fundamentals of chemical reaction engineering to understand the system, formulate the mathematics, so that the mathematics is able to tell us how we can manage our systems better than that we do now. That is the object and therefore it is a very core problem in reaction engineering, therefore it is we are not going away from the fundamental mandate of chemical reaction engineering. Having said this, a few points that I would like to draw your attention is that in our formulation, we said there is something called BOD. We said S10 is the BOD of the river as soon as it receives pollution. What are we saying? What we are saying is that the river has water and this water gets mixed up with pollutants and as a result, the river has a concentration of pollutants which is given as S10. How do we calculate S10? You know the flow of the river, you know the flow of the pollution, therefore you can do a material balance and find out what is the pollution level. Now, when you are trying to measure pollution, we said there are two kinds of measurements that we normally do. What is that measurement? One is we do BOD, one is we do COD, third is we measure oxygen concentration or so called dissolved oxygen. Now, BOD is a measurement which takes five days. BOD is a measurement that takes five days. That means, whenever we take a sample and determine BOD, it is at the end of five days. That means, you put us, you take a sample, you enrich it with oxygen. Generally, this enrichment is done by putting good water which is saturated with oxygen and then you wait for five days. At the end of five days, you find out how much of the oxygen that you have put in is still remaining. The difference between initial oxygen level and the present oxygen level at the end of five days is called as the amount of oxygen that is consumed and then you relate it to the amount of pollution water that you have charged into the vessel. So, this is how it is calculated in the laboratory. On other words, what we are saying is that BOD 5 is a measurement that you and I will do in a laboratory by taking a sample. But as far as the oxygen consumption of the organisms of the river, what matters is that the ultimate level of the BOD. We have done only five-day BOD, but actually what matters to the river is the ultimate BOD of the pollution. Therefore, given five-day BOD, we should calculate what is the ultimate BOD. How do we calculate ultimate BOD given five-day BOD? Now, there are various ways in which you can do this. I find it convenient to do an induction. What is induction? I say induction means that I take this polluted water and I put this polluted water into my water containing saturated with the oxygen. Now, that means at time t equal to 0, I have put my pollution into this flask containing good water which I am going to take to my laboratory. Now, at time t equal to 0, what happens? How much oxygen is present in that vessel? Very clearly, we are not allowed any time for the organisms to consume that oxygen. Therefore, at time t equal to 0, in your bottle where you have good water containing oxygen and bad water which consumes your oxygen, at time t equal to 0, no amount of oxygen would have been consumed. Therefore, t is 0 means what? t equal to 0 means what? The BOD that you will read at time t equal to 0 is 0. Is that clear? Let me say this once again. What is BOD test? BOD test is a test in which you take waste water, maybe 1 ml, 2 ml, whatever. Then you put this waste water into a flask containing good water which is saturated with oxygen. You know the total amount of oxygen that is in that flask and you put this 2 ml or 3 ml of your sewage and you close that flask. Now, what do you expect? As soon as this organisms present in this sewage of yours starts to consume the oxygen that is present in your flask which is closed. So, no more oxygen can come from outside. Now, you wait for 5 days and find out how much of oxygen that you have initially in the flask, how much has been consumed. Since you know the amount of oxygen consumed, you know amount of sewage that you have put in, you are able to determine the oxygen demand of your sewage. Now, what I am saying now is something else. What I am saying now is you have a beaker which is containing good water saturated with oxygen. You have put your sewage at time t equal to 0 and immediately took a sample and measured the amount of oxygen that is present. You will find no difference or in other words when you allow only 0 time for the for the sewage to consume the oxygen, you will find that the BOD that you will measure is 0, correct. Because you have not allowed any time for your sewage organisms to consume the oxygen of your flask. Now, if you allow infinite time, it will consume a lot more oxygen, correct. On other words, by induction, if you look at this formula here, if you look at this formula here, when time t equal to 0, e to the power of 0 is 1, 1 minus 1 is 0, therefore at time t equal to 0, BOD 0 time is 0. BOD 0 time is 0 because you have not allowed any time at all for the organism to consume the oxygen of your vessel. Therefore, BOD infinite time will be very, that will give you what? BOD infinity, put t equal to infinity, it will be BODL. So, it will give you BODL. On other words, what we are trying to say is that, now the BOD that you will measure at the end of 5 days divided by 1 minus e to the power of minus k 1 t, where t is 5 is the ultimate BOD of the sample that you have charged into your equipment. Let me repeat, what we are saying is, if you want to find the ultimate BOD of your sample, you will measure 5 day BOD of your sample, then the ultimate BOD of your sample is 5 day BOD divided by 1 minus e to the power of minus of k 1 multiplied by 5, where k 1 is what? k 1 is the rate constant for BOD removal and that number you know through an independent experiment, 0.7 per day may be or 0.5 per day if it is a cold climate and so on. So, what we are saying now is that, once you know BOD 5, you will determine BOD ultimate using this formula. So, when you are trying to determine the effect of the sewage that is entering the river, what is of concern to the river is the ultimate BOD of your sample. You are not able to determine the ultimate COD of a sample, because your experiment goes for only 5 days, we are all very busy, we cannot wait forever, we have to do it quickly. Some people do only 3 day BOD nowadays, in which case this t here is 3 days. So, whether you do 5 day BOD or 3 day BOD, the point is that BOD ultimate is related to BOD 5 by divided by 1 minus e to the power of minus of k 1 multiplied by 5 days, if it is a 5 day BOD, if it is 3 day BOD it is 3 here. So, you have now determined the BOD of the sample of the river that you have to which pollution is entering. Now, let us see what happens. Now, what we have tried to say here is that after all you see you have in your model, you have beta, you have k 3, your various items that appears in your model. I would like you to understand the meaning of all these terms. What is beta? We said beta is the oxygen load coming from the natural biology of water. What is the natural biology of water? There might be sediments. There are organisms which will respire and there are photosynthetic organism which generates oxygen. On other words, the natural biology of water that beta is something it is the oxygen that is used up by the natural biology. By and large, we should not interfere with the natural biology of water. On other words, the oxygen that is naturally available to that water, we should not interfere. The oxygen that we might be able to make use of is the oxygen that is coming from re-aeration k 3. I mean, even this there are many environmentalist who feel that even this is not for us. We should not interfere with this also. But the oxygen side literature suggests that part of k 3 might still be usable. What is the meaning of s critical? s critical means it is the value of s at the point of maximum sag. We pointed out just now that when you put pollution into the river, there is a point of the river somewhere downstream where the oxygen becomes very low or the sag becomes very high. That is the worst point of the river from the point of view of the health of the river. Our interest is to see that the worst scenario of the river should still be acceptable to as per our norms. That is what we are trying to say. The worst scenario should still be acceptable. That is why we want to know what is the worst scenario. For example, if the s critical, s critical, what is that means the pollution load in the river at s critical at the point of maximum sag. Let us say this BOD here is 20. Then this is unacceptable because it says that drinking water source should preferably have a BOD less than 3. BOD less than 3 is considered to be a drinking water source which we will take, we will purify, we will treat, we will remove all the BOD, we will sort of restore that water quality to quality that is required for consumption by humans and our domestic animals. So, that is the meaning of s critical. The meaning of s10, what is the meaning of s10? s10, you have this pollution entering the river. River has a certain amount of flow. The two will mix. You can do a material balance and find out what is the pollution load after having been mixed up with the river. The point we are trying to say is that river has a certain flow. Your waste is coming at a certain flow. So, when they mix, the mix pollution load is what is the interest as far as the self-cleaning power of the river. That is what we are trying to say. So, most important point that we must bear in mind the whole object of doing this oxygen sag analysis is to determine what is the worst point in the river from the point of view of oxygen sag. Then that worst point be such that it is not undesirable to draw water from that point for drinking purposes. See, the worst point also we must have a BOD less than 3. Then only it is suitable for drawing water from that point for drinking purposes which we will treat and then bring it to appropriate norms and then give it to people. Now, if it so happens that point is very bad, clearly we must tell the concerned municipalities please draw water from that point because that water is in very bad shape. That is another way of I mean telling the municipal authorities that listen this point is unsatisfactory. Therefore, we must go downstream so that the water quality is better. So, it is from that point of view of understanding the fundamentals of the reaction engineering that is required to understand what is the status today to understand what you and I can do to restore that status. This is the object of what we are trying to do having said this we want to do a few examples to quickly get to get a feel for what these numbers are. And I have taken examples from different parts of the world to tell you that situations in India are very very different from situations in particularly the western world. The western world is there is water abundant. So, it is plenty of water in the river there and dissolved oxygen in the river you can see here the dissolved oxygen in the river is about 9 milligrams per liter because temperatures are low and look at the river has huge amount of flow huge amount of flow. We do not have this kind of flow even during monsoons in many of our rivers you see. So, I mean I am doing this just to draw attention to the fact that you know we must look at our systems very differently and to be able to do it differently we must appreciate what happens in other parts of the world. So, that you know we take our lessons from them you see what happens in the western world Europe and America versus abundant amount of water. So, even if you put sewage into the river that is the example we are doing now it is not very serious as you will see right now. The first example we have taken here is that there is a river with 10 cubic meters per second of flow where the duo of the river is 9 milligrams per liter and then the sewage is entering at 30 million gallons per day and it coming with the BOD 5 of 200 milligrams per liter. It is coming with 200 milligrams per liter BOD 5 now how do you calculate the mixture duo of the mixture duo of the mixture is hot the 9 is what is coming in this sewage has no oxygen it is absolutely no oxygen in this therefore total amount of oxygen in the river divided by total amount of flow that is coming in plus the flow that we are putting in that is what I have written here 10 multiplied by 3600 multiplied by 24 this is the flow in the river 13 to 3800 is the flow of sewage and on the numerator is the total amount of oxygen that is contained in the water of the river which is 9 milligrams per liter and then 10 which is cubic meters per second 3600 per hour per day. When you do all this calculation we find that this oxygen in concentration in the river is 7.95 what does it mean 7.95 itself is a very fairly good concentration it is a good water as far as aquatic life is concerned and this is what I want you to appreciate that is because there is so much flow in the river they can do this in fact much of the sanitary engineering literature the books are written from the western world and these kinds of numbers are not uncommon you see. Now suppose I ask you what is the BOD of the river after the after it is got mixed up with sewage once again what I have done so much of BOD see 30 30 MGT I have multiplied by 3800 1 gallon is 3.8 liter that is why I have multiplied by 3800 and it is coming with 200 and the the river is coming with the BOD of 1 the river BOD of river BOD river is 1 1 milligram per liter that is I have taken that also I have taken that. Now you can calculate all this and you find that the BOD 5 of the river is 24.1 as soon as it mixes as soon as it mixes the BOD 5 is 24.1 the river flow velocities are all given so it is river flow is 3 kilometers per hour to cut the long story short now what is the question that we want to answer the question we want to answer is this BOD 5 which is 24.1 this dissolved oxygen which is 7.95 what is the what is the oxygen sag at this point the device 9 9 minus 7.95 it is 1.05 that means psi where oxygen sag is 1.05. So what I have done I have done this calculations because they are not very difficult to do what I have done I have calculated the ultimate BOD what is the ultimate BOD is 31 how do you find ultimate BOD 5 day BOD divided by 1 minus e to the power of minus of 0.3 into 5 what is 0.3 0.3 is k1 that is the rate constant for consumption of BOD that is the data is given the lot of data is available it is 0.3 because the cold climate and this rate constants are quite low okay. So it gives you 31 the initial deficit is calculated as 1.05 what is the re-aration constant is 0.7 now notice here that the re-aration re-aration constant of 0.7 if you go to cold climates like Canada and the US and the lot of ambient turbulence the velocities are such that it is a plenty of re-aration okay the temperatures are low okay oxygen solubilities are high so there is a benefit of oxygen solubility the driving forces are favorable for oxygen supply. So k3 is 0.7 k1 is 0.3 therefore you can calculate what is tau max where it occurs I found out it occurs at 1.96 days that means suppose you put the pollution here at this point 1.96 days later only you will the oxygen sag becomes the highest if the flow is if the river is traveling at 3 kilometers per hour you can calculate what is the distance it will travel may be 100 kilometers whatever so what we are saying what we are saying is that as per the model that we have set up that when you put 3 30 mg d of pollution 30 mg d of pollution into this river having so much of flow then you find that the maximum sag occurs at 1.96 days 1.96 days and at that 1.96 days what is the value of s critical s that means what is the value of BOD at the point why the oxygen sag is the highest then it is given by an exponential decay in 1.96 days it becomes 17.2 what we have said what we have said is that you have started with s value of 31 it has become 17.2 okay oxygen sag it started with 1.05 the maximum sag what is the maximum sag I would have calculated the maximum sag say I have calculated the maximum sag here as 7.29 what we are trying to say here is here is an instance of a river where the maximum sag occurs at about 1.96 days afterwards 1.96 days means about 140 kilometers these are all huge rivers this is huge rivers 140 kilometers afterwards the maximum sag occurs okay and plenty of flow in the river and then the maximum sag itself is 7.29 showing that even at the point of maximum sag the oxygen concentration is about 1.7 mg per liter. So, what we are trying to say the rivers of cold climates have plenty of water and even if you put the sewage in those rivers the sag is not it is not it is not very satisfactory but it is not like what happens I will show you in Indian rivers what happens this is not as bad as you would have thought it is still suitable for many types of fish okay that is the point we are trying to say. The second point a related point which is also of concern to us which I want to draw your attention is the following that is where are we here here now we said that at the point of maximum sag the maximum sag is 7.29 we have already said how did we calculate maximum sag we have a equation for maximum sag we just as put all the numbers I put all the numbers here you get maximum sag is 7.29 see at the point of maximum sag what is the value of s the point of maximum sag the value of s is given by 17.2. So, what we are trying to say here is that 17.2 mg per liter is not acceptable from the point of view of drawing water for drinking purposes. So, what we are saying is that even in cold climates where there is abundant amount of water in the river if you are putting so much of 30 million gallons of sewage into the river that you find that 140 kilometers later even 140 kilometers later that water is unsuitable for being drawn and then to be treated for purification that water itself is seen as unsuitable because this specification is that the water quality must have a BoD less than 3 that is from that point only you should draw your water that is the point I am trying to say. So, even in the case of cold climates like the western world the putting sewage into the river is creating some difficulties as you can see from the data. Now, let us go to the next exercise which is of direct interest to us that is suppose instead let us look at a river let us look at a river which is not cold climate but it is a warm climate. Let us like for example in South America you see South America is water surplus continent plenty of water when you go from northern north from the river Orinac to Amazon to you know Laplata is full of water plenty of water. But even there I want you to understand and appreciate that since the average temperatures are little larger temperatures are not as cold as in the United States in Canada and Europe temperatures are larger and as a result your K 1 is 0.5 and not 0.3 K 1 this is the one that you know this is the BoD removal rate constant it consumes oxygen you see it consumes oxygen therefore this is 0.5 when you do your calculations we will not do through this once again. But if when you do your calculations you recognize that you know you can go through the same I do not do it again you notice that when you go through this whole calculations you find you find that things are a little worse than what we saw in the cold climates that that the initial deficit previously was 1.7 now it is you can see here that the things are little worse than what we said before. As a result as a result you will find that the maximum sag is more than what we had thought in the cold climates to go through this once again what we are saying is that when when you have a cold climate and a warm climate when it is warm climate you find here is a BoD we started with 24.1 but BoD 5 is only 26 because it is 0.5 the rate constant is higher. So that is why previously it was 31 now it is only 26. So you find that the effect of a higher temperature it is important to appreciate these numbers when the temperature is higher the things behave differently and we must take this into account in our design. I have taken these numbers for Latin America because in Latin America there is plenty of water plenty of water and therefore the situations are not I mean so bad because there is so much of dilution due to so much of water. Now what what I want to draw your attention is the following even though there is so much of water so much of dilution the fact that the temperatures are high you will find an interesting result which I have calculated here. What is the interesting result the interesting result is that now you will find that the after after some time after some distance the the because the saturation solubilities previously the saturation solubility at cold climates the saturation solubility was very high close to 10 milligrams per litre in a warm climate the saturation solubility or 66.5 very warm climates because the saturation solubilities are low you will find that if you put so much of sewage into the river river will very quickly you will find that it becomes the oxygen sag becomes very high 6 oxygen sag becomes 6 but 6 means what it is all that means the DO is 0 because the CS the solubilities are very very low it is not 10 milligrams per litre it is much less. So what I am trying to say here is that when the temperatures are high when the temperatures are high you will have these kinds of situations where you will find for some distance for some distance the oxygen level in the river will be very low very low in fact nil and it will take some distance before it recovers on other words it recovers only after about 11 days. So what we are trying to say here is that in warm climates even in situations where the river has plenty of water even situations where river has plenty of water if the solubility of oxygen is low you run into various kinds of problems of oxygen level in the river and that will lead to various kinds of problems about health of the aquatic population what I am trying to put across to you here is that even in water surplus regions like latin America these problems can become quite serious we talked about cold climates there also the problems were of course I would not say it was very bad but it is it is serious if you put river sewage into the river if you go to warm climates if you put sewage it becomes it can become very serious now I am looking at a situation where the river has very little water as it happens in our case our rainfall is only for 2 maximum 30 days in a year that is it therefore our rivers are practically no water now when you put sewage into that river what happens this is what has been done here. So we have gone through the same calculations but this time we are doing for a situation of a river where the river itself is coming with some COD why is that the river is coming with COD is because there is so much of interference of the river systems by human activities the river itself is not in great shape if you go to western rivers in the western world the COD of that river will be 1.5 in our case it will be 10 that is when you do not put sewage into the river you put sewage into the river things are very different. Now we have done calculations here for the case of a river receiving essentially COD when the river is receiving essentially COD the reason why I am doing this for you is that when you are putting COD into the river the rate constants are very low COD is a molecule which is very difficult to get digested consumed by the bacteriology of the water. So you do not get 0.5 and all that even in warm climates like ours the rate constant may be 0.2 0.15 things like that I have taken 0.3 here but I want to draw your attention to this fact that COD is a recalcitrant molecule it is very difficult to digest there are only some bacteria which is able to manage very very toxic chemicals. So we have done calculation for that case we have taken here this is S2 and here it is K2 we can see here that means you have put S2 which is COD K2 is the rate constant for COD removal you have gone through all the calculations and once again here we have said the amount of pollution that is coming in is 100 KLD the amount of water in the river is only 0.1 cubic meters per second 0.1 cubic meters per second is not small it is still 360 cubic meters per hour 360 cubic meters per hour which is not small you see and even this is the kind of river we actually our rivers do not have much flow at all even this amount this amount of water might be there in Ganga but point is that Ganga also has huge amount of you know sewage flows from populations and so I have done a calculations it just put taken some numbers just to draw attention to the fact what is the kind of you know river behavior you will see and then the numbers I have gone through all the calculations let us not spend too much time I have calculated COD I have calculated the oxygen deficits what is the deficit and so on okay we are starting well it is 0.66 okay now you notice when you do all these calculations you notice the same problem that I have said before what is the problem the problem is that the the the S criticals are very large S criticals are very large it is unacceptable okay alright so what is the what is it that we are trying to say what we are trying to say that whether it is reverse of the western world whether it is reverse of the I mean Africans or in South American continent when it comes to reverse of the Indian subcontinent since the river flows are very low the river is just not having the enough capacity to manage pollution okay and therefore we are really not able to draw water from the river for supplying water to our population because that water itself is not in good shape okay so that is why we have to ensure that we do something about this whole problem okay now it is a related a related issue I have considered here a related issue I have considered here which is important let us also not forget that you see what what actually happens is that our waste waters contains both COD and and COD and both are there that means our situation that we must handle is that there is both COD and COD which we have to take care on other words our solution to the problem should actually treat both the both the cases and I have done that I am just please I am not showing you the details because the valley straight forward integrations so when you integrate the the whole situation this is what you find what you find here is that there is effect of COD there is effect of COD there is effect of initial deficit there is effect due to natural biology COD COD effect COD effect initial deficit effect and then the effect due to natural biology all the effects are present in the river that is what is the situation that you and I will have to content with okay now the last example I have one more example where I have taken both COD and COD into account and then I have taken some flows still this 0.1 cubic meters per second is still very large many of a rivers are not satisfactory in terms of these kinds of flows but I have done some calculations but I am sure you know once you have these numbers you can do these calculations yourself I have just these are some illustrative examples to draw attention to the fact that putting waste sewage or waste water industrial waste water even if it is treated waste water into the river is absolutely unacceptable because they just simply destroy the biology of the river okay that is the point I am trying to draw your attention that really speaking there is no oxygen capacity in the river to manage human generated pollution okay I have done these calculations I have done these calculations here and the net result the net result is saying the same thing that you know our rivers do not have any capacity to manage our pollution in fact the kind of answers that the kind of calculations that let me just go through this now that I have done these calculations I have done calculation for the case of a river with 0.1 cubic meters per second so you get a BoD of 4.6 okay that is the ultimate BoD of 4.6 and then COD comes out to be about 15.6 therefore this the COD level that we have to content with this 11 and BoD level you have to do is 4.6 and you put all these things into our equations you find we find that the same kind of behavior we see what we find is that for a very large section for a very very large section you know the river is like this you know this is the saturation there is a large region where the oxygen is very very 0 you see something that we are unacceptable okay as you can see here now you have to go up to about 6 you know 6 days of residence time before the river becomes a little better okay. So cut the long story short what I am trying to put across to you is that the tropical river systems of this country the flows are so small that they do not have any capacity to manage pollution load from human activities therefore to actually accept affluent norms of 250 milligrams per liter COD and all that it is something that 250 milligrams per liter COD if you let out into the system into the environment where where does it go it will go into the river or into the lake or into a ocean none of these systems have the oxygen that is required to manage all this. So let us resolve and to recognize that we must design treatment systems which are all reaction equipment all treatment plans are reaction equipments in which we must remove the pollution level to such levels that our natural systems can handle this pollution okay which means COD less than 10 I would say COD less than 5 BOD less than 2 DO at least be close to saturation this is what we should let out so that our river systems our lakes our rivers our oceans are in good shape.