 in today's class, we will discuss about reservoirs. And when we say reservoirs, the purpose may be manifold, it may be used for flood control, it may be for irrigation, it may be for power generation, it may be fishy culture or aqua culture, it may be drinking water supply, it can be sometimes recreation. Just remember that we may build a reservoir due to one or more than one of these reasons. So, suppose a channel is flowing with water, we should know what is the inflow to the system and then accordingly, we can design the reservoir based on our requirement. When we say requirement, also sometimes we call it the demand. So, two terms, one is the demand, the other is the supply and based on these two, we will design the reservoir. So, when we say reservoir, our purpose is basically threefold. One is to calculate what is the demand, then what is the reservoir capacity and then how we should operate the reservoir. We will discuss about these two and may be you will solve some problems. So, there is something called the water availability. What is water availability? Suppose I have the data for 50 years, let us say the Delhi discharges for a stream for 50 years. So, what I should do is accumulate these values for month or calculate monthly discharges. That means, you sum the Delhi discharge values for the whole month, then now you get 12 values for a year. So, for 50 years you get that many values, 50 multiplied by 12. Now, suppose I want to know what is the water availability. So, I can plot here, this is 0 percent, this is 100 percent and this is discharge and the series I get, I can divide them in descending order. That means, the highest should be at the top, you can say z to a. Now, when I plot these values something like this, the highest value I should assign 0 percent and to the lowest value I should assign 100 percent. That means, 100 percent of the time this discharge is available to me. So, suppose I want to find out what is 75 percent dependable flow, then go here, this is 75 percent. Suppose I want to know what is 90 percent, then I go here and find out what is the discharge. So, this is 90 percent dependable flow, this is 75 percent dependable flow. A thumb rule for deciding the reservoir is generally 75 percent dependable flow is used for irrigation and 90 percent dependable flow is used for power supply and 100 percent dependable flow is used for drinking water supply. Now, let us go to the next thing to decide how I can find out the reservoir capacity. As I said earlier, we need to know what is the demand or what is the water requirement. The purpose can be manifold, whatever is the purpose depending on that, what is the requirement of water and what is the supply of water. Now, I will introduce you two terms, one is the mass curve and later I will tell you about the differential mass curve, let us see what is mass curve. In mass curve, I plot here time and in the other axis I plot cumulative volume, that means the curve will look like this, because this is cumulative volume. So, a value let us say here, it cannot be less than it is previous value, because what is cumulative? If I have a series, suppose this is t del e or 10 del e or monthly or yearly values and I have here the flow volume. Suppose, this is time period 1, 2, 3, 4 like this and here I have let us say 100, 110, 80, 30, may be 120 like this. Then this cumulative volume will be this will be this 100 and then I keep on adding these numbers, that means 210, then 80, 290, then 30, 320 like this. So, subsequent values or in this series a value cannot be less than the previous values. So, this is called the mass curve. So, for this stream under consideration for which we are trying to design the reserve here, for that reserve here we need to find out what is the mass curve of supply. So, we find out we calculate from the given data mass curve supply and remember this is mass curve for flow volume and also we need to prepare mass curve for demand and on the same plot we try to see these two curves. Let us say supply is like this and demand is let us say like this. So, in this case this is supply, this is demand. Then we know that at this place supply is more than demand. Let me make another curve with dotted line so that it will be more clear. This is time, this is mass curve or I can say q it can be supply or demand. Let us say supply is through a solid line. So, supply is like this. So, this is s and let us say demand is a dotted line. So, this is demand. So, consider this portion from this to this. Let us say this supply curve is like this. So, here supply is more than demand. We do not require the reserve volume to store water because supply is already above demand, but consider this portion. Here demand is more than the supply and we need some reservoir. We need some storage so that we can store the water and whenever required at this time from this time onwards we can supply that water from the reservoir. This is the supply you do flow of the stream, but we should reserve water so that we can fulfill this demand. Let us see a systematic procedure to find out what is the reservoir capacity. So, first of all you prepare the mass curve for supply and then you prepare the demand curve. Let us say this is like this and wherever demand is higher you should concentrate on that location. For example, here demand is more. So, I should find out what is the difference. Similarly, here demand is more. I should find out what is the difference. Similarly, here demand is more. I should find out what is the difference. Here demand is more. I should find out what is the difference. Let us say we have 1, 2, 3, 4 values. At four different places the demand is higher than the supply and we need to know what is the extra amount we should store to fulfill the demand. So, I should find out maximum of these differences. Let us say this is sigma q demand minus sigma q supply. So, wherever you consider this all these times and wherever this value is the maximum that will be the reservoir capacity. This thing will be more clear when we solve one problem and let us go to the next thing. Just now I indicated that demand is through a constant curve like this. Do you think the demand will be constant? What does this mean? This means in time this is time. Let us say these are the values against each month. So, each month the demand is constant. That is why this is a straight line with constant slope because when I keep on adding let us say this is x, x, x, x, x always then when I keep on adding this will give me a straight line and with a constant slope. So, sigma q will be x 2x 3x 4x like this. So, we get a straight line. But do you think this is in reality this is the case? No. What we observe is the demand is also a variable in time. For example, during summer we need more water. Similarly, during winter we may need less water. So, this demand will again be a variable thing. Let us say this is the demand variable demand and let us say similarly this is the variable supply. Then I should consider different time at different time. What is the difference? Out of these differences I should accept or my design value or the reservoir capacity should be the maximum one. So, I consider the maximum of these differences. So, whether it is a constant demand curve or a variable demand curve it does not matter. What we have to do is plot the mass curve both for the demand and the supply and try to see where is the maximum difference. Based on that maximum difference we plot, we decide the reservoir capacity. Now, I will pass on to the next thing which is another technique to find out the reservoir capacity it is called the differential mass curve. The previous thing is called the differential mass curve because we plotted the mass curve and then we find out the difference. Now, this technique which I am going to tell is called the sequent peak algorithm. So, in the sequent peak algorithm generally a computer program is used and please remember that to design the reservoir to arrive at a number for the reservoir capacity what you do is it is not just 1 or 2 years. Generally we consider for the proposed site we consider 40 to 50 years discharge value. So, when we consider such a long series you know better to use a computer program. So, the sequent peak algorithm is a computer program it is an algorithm which very systematically calculate the reservoir capacity and let us see that procedure now. In sequent peak algorithm what we do is this is let us say time and this is let us say supply and this is demand. So, what we do is in the previous method in the differential mass curve first we find out the mass curve for this and then mass curve for this we compare the mass curves then find out what is the capacity, but here what we do is first we find out the difference. So, we find out the difference you can say difference between these two columns. So, you get numbers here. So, suppose this is supply this is demand then you get a series some numbers will be positive some numbers will be negative you get a series like this. Remember if supply is more than demand then supply minus demand will be positive if supply is less then supply minus demand will be negative. So, in sequent peak algorithm what we do is we try to find out the peaks suppose this is a peak remember this is time and what is this in y axis I have the fourth column in the previous table I showed this fourth column this is time this is supply this is demand and this is difference of these two difference of these two. So, this column all right. So, this is that difference. So, what I do here is I consider the peak and then the next peak let us say this and I neglect these two peaks because these values are less compared to this peak value because this is called a sequent peak algorithm. So, I have to search the next higher peak. So, let us say this is another peak and if the series continues then I should find out where is the peak and remember I do not consider these peaks because these peaks are less compared to this peak. So, once I am here I should search for the next higher peak compared to this peak. So, in this particular figure I have 1, 2, 3 may be 4, 4 peaks. Now in between the peaks let us say this is peak 1, this is peak 2, this is peak 3, this is peak 4 and remember the decision how to find out the peaks the peak has to be higher than this value and these two I should not consider because these are less compared to this. So, here in between two peaks I should find out what is the minimum trough. So, suppose the 1, 2, 3, 4 troughs are there out of these four troughs I should decide the minimum one. Similarly, in between these two peaks there may be more than one trough or minimas I should find out the minimum one. Similarly, in between these two peaks I have to find out whichever is the minimum value. Now let us say this is T 1, this is T 2, this is T 3. Now for calculating the reservoir capacity what I should do is you just find out P 1 minus T 1 for the difference between these two. Similarly, P 2 minus T 2 and P 3 minus T 3 and find out what is the maximum between these three. Do not forget that in actuality in the series it may be a very long series and in the series your computer program will do these operations and it will try to peak the numbers the peaks and the troughs and in between the peaks and troughs it will try to find out what is the maximum where is the maximum difference depending on that the reservoir capacity will be decided. Now I will repeat once again in these differences while calculating these differences you have to be slightly careful because this will be with a minus sign and this will be with a plus sign. So, when we say difference basically we add these two magnitudes this is the peak this is the trough. So, we add these magnitudes to find out the reservoir capacity and these things will be more clear when we will solve some problems. So far we have discussed about the estimation of reservoir capacity and we have discussed about three different things one is the constant demand the other is the variable demand and also we have discussed about the algorithm which is sequent peak. Now also you should remember that conversely one can do the problem of suppose a reservoir capacity is known the reservoir volume is known then can we determine the volume the water requirement what is the demand to be fulfilled by this reservoir can we determine that yes. So, now we will see different problems so that these concepts will be more clear. This is problem one calculate the minimum storage required to maintain a demand rate of 40 meter cube per second and these are the flows given for 12 different months in January 60 Q mech please remember these two units Q mech and meter cube per second are same. So, these are the numbers given different inflows for a year and what we should do we should first prepare the mass curve for flow when I say flow that means this is the supply then I should prepare the mass curve for demand and I must compare these two to find out the difference that is the reservoir capacity do not forget that mass curve is time this is cumulative flow. So, here are the steps for different months I have the supply so this number 60 indicates the average Delhi discharge. So, I should multiply 31 because January has 31 days so this is the monthly value the 60 is the average Delhi discharge and for the month this is the discharge you can say this is the supply. Similarly, for other months we have these Q mechs or meter cube per second as supply then this is the demand this is a case of constant demand. So, the water required is 40 Q mechs again these are Delhi values. So, for respective months I should find out the number the monthly values so this is again in Q mech. So, now what should I do I should prepare the mass curve remember these numbers are only for the month. So, I must prepare the mass curve then only I should compare what is supply what is demand and what is the required reservoir capacity. So, this is the mass curve for supply remember the previous numbers. So, what should I do I should add the first number will be 1860 next I should add this to this and similarly I keep on adding. So, I get the cumulative curve which give me the mass curve for supply this is the mass curve for supply. So, in x axis I will have the months here I will have the months from January to December and in this axis I will have the supply. So, whatever will be this whatever number I get this will be the mass curve and similarly for demand I can get the mass curve by adding the previous column values. Now, I must plot this and this let us say this is something like this. So, here it will not be like this because the demand at some places will be more than the supply then only we require a reservoir if the supply is higher than the demand then there is no need for the reservoir. So, let us say this is something like this then let me draw here one more figure let us say this is the demand because we have a constant demand. So, it will be with a constant slope and let us say the supply initially for example here 1860 is more than 1240. So, initially it is more but then gradually it will be less than that and then towards the end it is again more the supply is more. So, if you examine the critical months will be this is January, February, March. So, look at here. Here also I have more the supply is more compared to the demand but look at this. This is these three months are the dry seasons and these three months are critical and we need a reservoir to cater the requirement of water and these three periods will indicate what is the required capacity. So, if I plot these values I can straight away find out suppose I find out the difference here for example here the demand is 72040 and the supply is 6080. So, this is 1160 is required but you know sometimes these points let us say this is one point this is another point. So, the deep or the my intention is to find out the difference between these two plots at different points and for that matter what should I do? I should draw the transient at different locations parallel to this line here I may have to draw like this. So, depending on the maximum difference from the demand we determine the capacity and if you do this graphically we will find out this value will be slightly higher. Here the analytical number is 1160 but if you do it graphically you may find it to be slightly higher I believe the value will be around 1800 or 1900 will be the value that means the reservoir capacity will be that much. So, what I have to do is I plot this curve I plot this curve and take the difference and wherever is the difference is maximum I consider that to be the reservoir capacity. I think the case of constant demand is clear now let us see the second case this case is calculate the demand that can be fulfilled by a storage of 3600 cubic days. So, this is the case of reverse problem that means in the other problem demand was given and we have to find out the reservoir capacity here reservoir capacity is given and we have to find out what is the demand that can be satisfied through this reservoir. So, the exercise will be pretty similar these are remember these are the I will come back to this once again these are the inflow values. So, the procedure will be like this I have to prepare the mass curve for the supply this is supply. So, prepare the mass curve for the supply and now you do not know the demand, but you know the reservoir capacity. So, what you have to do you guess some lower values in the mass curve suppose this is the mass curve for the supply. So, maybe you can take this or this or this you guess some time. So, there you draw the tangents then let us say my first gas value is this I draw a tangent here then add storage I know the storage in the problem storage is given. So, add the storage. So, here now you know to this you can through this you pass a line like this because you assume that the demand is still constant. So, this mass curve for demand will have a constant slope and if I get this line now what is to be done I know that this is the mass curve for demand. Now, once this curve is known I can easily find out the monthly values. For example, here it will be with a constant value you can deduct the previous months value for example, here the monthly value will be this minus this. So, you get the demand discharge monthly demand discharge. Now, the objective is in this problem to solve the problem for the demand curve. So, we know in the reservoir capacity and through this exercise through this problem we can know what is the demand that can be fulfilled through this reservoir. Now, let us see the next problem which will be on variable demand. Please remember that for the sake of simplicity for the sake of explaining you the things I assume that the demand is constant it need not be. So, let us see the next problem. So, this is for the previous problem for a proposed reservoir following data here calculated data will be given next. The prior water rights required the release of natural flow of or 5 cubic whichever is less this is important to understand. Sometimes suppose you build a reservoir but there might be some rulings you have to maintain some flow you cannot tap all the water. So, here in this question it is given that the release of natural flow 5 cubic whichever is less you have to release that flow for the requirement. This is remember this is not the demand this is the water rights requirement. Then assuming an average reservoir area of 20 kilometer square the reservoir area is given estimate the storage required to meet these demands. So, let us see the data this is the data these are months again the data is for one year. The second column is flow means flow in the stream and then this is the demand and this is the evaporation. Now, when water flows there will be some loss due to evaporation. So, this is evaporation and these are the rain falls. We have to be very careful because here in this particular question by mistake no units are given here. So, in the data you will be provided with units for example rainfall may be in centimeter or in millimeter. Similarly evaporation will be in terms of some depth and flow will be in cubic and this demand will be again in cubic. So, these are the data. So, what is the procedure we have to follow same procedure consider all inflows and we call it a supply. So, for the system for the given system you consider all the inflows and what will be the inflows? Let us see this will be inflow this will be positive this will be also because rainfall will add to the system. So, rainfall should be also considered as inflow to the system. Now, let us see what are the demands consider all outflows because demand is something which you have to provide the reservoir has to cater. So, this while calculating demand we must consider this is the water required and also evaporation because this is a loss for the system this is a loss. So, these two column will give me the positive things for the inflow to the system these two columns will give me the negative things or loss or from the system. So, you now prepare the mass curves as usual that means in the x axis you take time or months in y axis you take cumulative flow. Here there is a trick you should be consistent as far as your units are concerned. Here I will give you a clue the clue is either you use volume for all the components for example rainfall you should consider volume or you should consider depth as the unit. So, you are free to use either volume or depth as the unit. So, while preparing the mass flow curve for the supply and for the demand you should be vigilant that your unit should be consistent. So, if you are adding for example here when you add flow this is in cubic this is in centimeter you try to write in one unit. If you are using volume then to this depth you multiply the reservoir area. So, you get the volume similarly if you are interested to find out the depth for this flow then you divide by the area. Now these two steps are over and I also know how to prepare the mass curve. Now it is all same compare both and as usual the maximum difference between the supply and the demand curve will give us the reservoir capacity. The only difference in this problem is the demand is not a straight line in the earlier problems demand was plotted like this because we assume that demand is constant. But here it will not be constant because of variable demands. So, it does not matter you plot your supply whatever is the supply and you plot whatever is the demand. And wherever it is maximum you consider at all places wherever it is maximum let us say this place it is maximum. So, that will be the reservoir capacity I think this is clear the example explains you how to take into account the variable demand. Now let us see the next one which is on sequent peak algorithm. The average monthly flows into a reservoir in a period of 2 years is given below. Please remember while you are planning for a reservoir suppose you are asked to design a reservoir you are asked to find out the reservoir capacity. Then the data should be for more than 10 years remember this only for the sake of simplicity we plotted here the data for 1 year. But in reality you have to use more than 10 years data and again in sequent peak algorithm it is very useful because you have a long series time data. This is the average monthly flows into a reservoir in a period of 2 years is given below. If a uniform discharge at 90 cubic is desired from this reservoir calculate the minimum storage capacity of the reservoir. Again here the demand is constant and you can also solve this problem the way we did the first problem which is through differential mass curve. But here we are trying to solve this problem through sequent peak algorithm. So, for June it is 20 July 60 and likewise we have here data for 1 year and next we have data for another 1 year. So, what is to be done we have data for 2 years the method says you if you are with n periods data prepare the data for 2 n periods why because we assume that the same flow will continue for the next 2 years. For example, here we have data for 2 years and we must prepare data for 4 years then where from we will come the data for the next 2 years the same data we will use for 3rd and 4th year. That means we have 12 values for 1st year 12 values for 2nd year for 3rd year this value will be used and for the 4th year these values will be used. Now find net flow volume for each month. So, what you have to do is you do not have to prepare the mass curve in the other method we have to prepare the mass curve suppose I have data for 24 months or may be 48 months. So, use this 48 months to prepare the mass curve here we are not doing that remember this. So, we use monthly basis and we try to find out what is the difference. So, what we have to do is we are not preparing the mass curve what we will do is on monthly basis we have to see what is the difference between supply and demand. So, for each month that means in the data set we should consider row wise. So, here we have the data for inflow and in the question this is the 90 kumec for demand. So, what you have to prepare is so what we have to do is we should consider row wise and in a particular row we should this see what is the demand what is the supply and what is the difference. So, we have to prepare data like this. So, this will be may be serial number this will be time and in our case it will be months and the data for our case is from June onwards and remember we have 48 values. So, this will be 1 this will be 48 and we have supply for each month we have been given with a number. So, that number you have to write here and then demand. So, here you put 90 everywhere and then here you prepare s minus d this is supply this is demand. So, you prepare supply minus demand whenever it is running or in the so called not dry period we will have supply more than demand. So, it will be positive whatever number it comes it will be positive sometimes it may be negative. So, for each of these rows one has to find out what is the difference between these two and you should also remember the procedure we followed and the procedure is what find net flow volume for each month and it is s minus d. In the computer program in fact you use a subscript i for this and this i is the index for the respective month and for the respective month you find out what is the difference and when I say difference it may be positive or it may be negative s minus d and then find cumulative net flow volume. And then this column in this column you have to do something like a mass curve in the earlier technique differential mass curve what we did was preparing the mass curve here after supply you prepare the mass curve for supply after demand you prepare the mass curve for demand and then compare these two, but here individually you prepare what is the difference at each month then you prepare a cumulative series that means sigma s minus d i. So, here it will be only one here it will be this plus this here it will be this plus this like this you will get some numbers. Now, this will be treated to find out the peaks and the drops and as we discussed you will suppose you will get values like this. Now this is your first peak you should see where is the next higher peak because this is a peak this is a peak this is a peak, but these peaks are less compared to this, but this might be higher than this. So, this will be if this is peak 1 this is peak 2 similarly in this series remember only this is an example and in this example we have 48 rows, but in reality it will be more than that you can guess if you have 50 years data how big will be the series. So, the computer will search the computer program will search and then you find out where is the sequential peaks that means one peak and the next peak should be higher than the first peak. So, similarly you find out all the peaks and in between 2 peaks you will find the minimum of all the drops here we have this as a drop this as a drop this as a drop this as a drop this as a drop these two are also drops, but out of so many drops I will go for this one why because this is the minimum one then you decide p minus t if you have more than one peaks and then corresponding drops in the gap of 2 peaks you find out the minimum drop similarly between p 2 and p 3 there might be so many drops and you decide you go for the minimum drop. So, p 2 minus t 2 similarly p 3 minus t 3 and you consider all these numbers and the maximum number will be your reservoir capacity. So, once again I will repeat the procedure which says use the sequent peak algorithm that means find out where is p 1 p 2 or t 1 t 2 t 3 and then find the maximum of all the p minus t values. So, out of all the p minus t series you find out which is the maximum one and the maximum value gives you the reservoir capacity this is please remember this sequent peak algorithm is more useful when we have a longer series and the if we have less data then maybe we can go for the differential mass curve and if you compare the results then they give more or less same results we should get same results. If one is using the differential mass curve and the other one is using the sequent peak algorithm one should get same result, the sequent peak algorithm is less prone to error. So, generally it is preferred. So, I think with the help of these four examples the concept of estimating the reservoir capacity is clear and in the next lecture we will study the design of Culver.