 So, we have looked at the process of runoff and we have seen that out of the total precipitation some part is abstracted through interception, diffusion storage, evaporation, transpiration and infiltration and the rest goes as runoff. This runoff is the component in which we are very much interested as water sources engineer because we want to know how much water is available at a certain point in a river. So, that we can use it to supply water or so that we can design some flood protection works if the magnitude of the water available is very high. So, in order to look at this we have to analyze in the quantity of water which is available at a certain point and for that purpose we sometimes for gauging stations which measure the height of water or the depth of water in the river and correlate that with the discharge. So, that if we measure the depth or what is known as the stage then we can obtain the discharge from the stage discharge curve. Now, this gauging station may not be available at all rivers, but let us say that some river has a gauging station the recorded the gauging station is typically in terms of stage which is converted to discharge and a typical record will look like this. So, this is the year and this denotes the date or the day of the year. So, first represents the first January and in this case 365 will represent 31st December. So, for the whole year we will have data available for the discharge through the river and what is typically done is the discharge is averaged over a day. So, this is the daily mean flow passing at that point. So, if we have a catchment and there may be a river and then there is a gauging station here at point A then measuring the stage at point A we can convert that into discharge and for the daily discharge data we can obtain what is the mean daily discharge at that point A for different times of the year. So, on first January we can see that the mean discharge was 10.2 meter cube per second on second it was 11 and similarly we have data for all other day in at day 200 the discharge is quite high at 121 meter cube per second and similarly at the end of the year it is 17.1 meter cube per second. This record is available for a number of years for example, we have shown 5 year data 2002, 3 2004 of course, we will have 366 days. So, this corresponds to 31st of December and then again 2005 we have 365 days. So, we have a lot of data 5 times 365 plus 1. So, in order to analyze all this data we need to reduce the data in some form. So, that we can analyze it more quickly and accurately. So, for that purpose typically what is done is we divide it into range of flow. So, instead of taking individual values of the flow we can say that we will divide the discharge into certain ranges. For example, here we have divided them into discharges of 10 for high discharges and 5 for lower discharges. So, this represents the discharge range again in meter cube per second. So, this is the daily average discharge in meter cube per second and what we do is we look at each year how many days are there on which the average falls in this range. So, for example, 2001 there is no data which is more than 140 or in the range of 140 to 150 in 2002 on 4 days the mean daily flow was between 140 and 150. Now, when we talk about the range typically we would include the higher range. So, this will be included and 140 would be excluded. So, if the discharge is exactly 140 then it will go into this range if it is 140.001 then it will go into this range. So, similarly in all other ranges and for each year based on the available data we can obtain the number of times or number of days on which the discharge is within that range. So, for example, here we can see that 35 to 40 there a large number of days with that range and then finally, we would our aim is to obtain a flow duration curve which tells us that what is the probability that a particular discharge will be equaled or exceeded how much percentage of time. So, for that we need to obtain the cumulative or the total number of days for this 5 year period on which the discharge was within 140 to 150 or 130 to 140 and so on. So, we can see that total number of days out of these 5 year data there are 6 days in which this charges within the range of 140 to 150. Similarly, there are 26 days on which this charges within the range of 130 to 140 and then we would do a cumulative discharge. So, this will be 6 then 6 plus 26 will give us 32 and so on. So, this 6 represents number of days for which this charges higher than 140 and 32 represents number of days for which this charges higher than 130 because 130 to 140 is 26 and 140 to 150 is 6. So, total of these 2 will give us the number of days for which this charges more than 130 meter cube per second. Similarly, we do all this and ultimately this 1827 represents 5 year data. So, 5 into 365 1 actually it should be 1826. So, this represents the total number of cumulative number of days for which discharge is exceeded. And now we can do a frequency analysis to analyze the probability of exceedance. For example, we write Q as we have done in the previous slide and then we write the cumulative again this is from the previous slide. This probability p is m over n plus 1, where m represents the cumulative total and n represents the total number of records in this case 5 years. So, dividing this cumulative number of days by the total number of records plus 1, we get a probability of 0.33 percent for discharge to be within the range of 140 to 150. Similarly, we can look at this charge of 5 to 10 and the probability is 99.95 percent. So, this tells us that almost 100 percent of the time this charge will be more than 5 meter cube per second. So, if we look at the flow duration curve which expresses this data in graphical form, here we have a probability in percent for a given Q and this probability is of exceedance, which means that if we look at any particular Q for example, let us look at Q equal to 70. Then this gives us a probability almost 25 percent that a Q of 70 meter cube per second would be equal or exceeded or in other words, we have a value of let us say 140 here, which has a very small probability of exceedance and then we have a value of 5 here, which has almost 100 percent probability of exceedance. So, what it means is that if we go to the river and measure the discharge every discharge over a day for any day, there is a very high chance that it will be more than 5 meter cube per second and there is very little chance that it will be more than 140 meter cube per second. The average discharge throughout the day is likely to be smaller than 140 and if we look at this point, then the probability that discharge will be higher than 70 meter cube per second is about 25 percent. So, this gives us a flow duration curve and we can use this curve to obtain the reliability of certain discharge. For example, when we design let us say a power plant, we want some discharge which will be available, we can say let us say 85 percent of the time. So, we can look at this and say that our 85 percent dependable discharge is about 25 meter cube per second. So, if we design something based on design value of 25 meter cube per second, then it is likely that 85 percent of the time the discharge will be more than that. So, this gives us an idea about how much flow is available in a river. Now, based on this availability of flow, we can design let us say storage system which can store water and then supply it when the flow is smaller. So, for this purpose, we analyze it as what is known as a mass curve and mass curve is nothing but how much accumulated mass of water is available throughout the year and to analyze a mass curve, we can look at in terms of monthly rate. So, instead of analyzing daily data, it is better to combine data for a month and plot it versus average monthly rate. So, if we look at this, January let us say that we have certain point again, we have a catchment and then at point A, we are measuring the stage and therefore discharge and the mean monthly flow at the point A for different months is let us say obtained as these values. So, we can see that the mean flow rate is very high in September just after the month owns and it is still higher in August and October, but then again it decreases as we go over here and decreases as we go over here and January, February again there is a reasonable amount of 45 and 48 meter cube per second. So, this represents overall an average flow rate throughout the month and based on this, we can find out the volume. So, for example, in January there will be 31 days multiplied by 24 hours then 3600 seconds. So, that gives us a volume which we have shown here in terms of million meter cube. So, this value of 120.5 represents the volume of flow in the river in the month of January. So, for the entire month of January 120.5 million meter cube of water will pass through this point A. Similarly, for February, March and so on till December and of course, September has the highest amount of volume available for the entire month. Now, we can add them up and find out the cumulative volume. So, 120.5 is here then 120.5 plus 116.1 will give us 236.6 then 236.6 plus 101.8 will give us 338.4. So, in this way we can obtain an accumulated mass which gives us 1682.6 million meter cube for the entire year and based on this value, we can obtain an average rate throughout the year. If we have this much volume available, we can convert this into an average rate which in this case turns out to be 53.4 meter cube per second. So, this means that the entire water available to us at point A can be represented by an average rate of 53.4 meter cube per second throughout the year. So, this rate is the maximum supply rate which we can maintain in the river throughout the year, but again we can see here that the flow available is only 45. So, if we have to maintain a rate of 53.4, we must build some storage and so that during these years water can be taken out of the storage because the supply rate is less than the demand of 53.4. So, in order to analyze how much storage capacity we would need, we plot the mass curve as shown here. So, this is for different months, the cumulative volume, this is again in million cubic meters. So, this shows for the whole year, what is the accumulated volume for a certain time. So, in January, we have so much accumulated volume and at the end of December, we have so much accumulated volume which we have already seen as 1682.6. So, this value is 1682.6 and if we join these two points, this will give us the maximum demand which we can satisfy which can be met. Now, typically we would not use the demand as 53.4, this value is the maximum value, but let us say our demand is smaller than this. So, for a given demand, for example, let us take the demand as 45 meter cube per second. So, for this given demand, if we want to find out what is the storage required, then what we need to do is we need to draw the line which has a slope of 45 meter cube. So, this line will give us an idea about how much storage we would need for demand of 45 meter cube per second. And in this case, suppose this is the 45 meter cube per second line, we draw a line tangential to this curve and parallel to this and then the difference between these two will give us the storage in terms of million meter cube. So, if we look at the calculation, what we can do is in tabular form, we can find out what is the difference between the available mean flow rate in the river and the demand which we want to maintain. In this case, the demand is 45, in January the mean flow rate is also 45. So, there is no difference. In February, the availability is 48 meter cube per second while the demand is 45. So, we have a surplus of 3 meter cube per second. So, till this point, we do not need any storage. Now, this 3 meter cube surplus can be converted into a volume in terms of million cubic meters. So, again 3 meter cube per second into 20 meter cube 28 days, 24 hours, 3600 seconds will give us meter cube and then we divide it by 1 million to get this value in million cubic meters. Now, after February, if you look at the March data, in March we have a supply of only 38 cubic meter per second while the demand is 45. So, that means we have a deficit of 7 meter cube per second which translates into a deficit of 18.7 million cubic meter for the month of March. In the next month, again we have a deficit and so on till we reach July in which case again we get a surplus. So, these 4 deficit months, they are the critical months for the storage capacity and our storage should be sufficient to take care of these deficit values. And if we add all these volumes, we get minus 205.1 which really corresponds to this. In the graphical terms, this storage numerically would be equal to 205.1. So, this tells us that if we want to have a storage reservoir built at this point A, we can build a reservoir with the capacity of 205 million cubic meters and we would be able to satisfy a demand of 45 meter cube per second throughout the year. So, this is the required storage capacity, yes. Similarly, if we change the demand, let us say now we want to have a demand of 50 cubic meter per second. In this case, we can see that from January itself, we have a deficit and then up to December, August to December, we have surplus. So, in this case, following the same philosophy, we can add up all these volumes and we get 284.1. In graphical terms, this would simply mean that drawing a line at 50 and then parallel to that and taking an S, which would in this case turn out to be 284 million cubic meters. So, this procedure gives us the required storage for a given demand. Now, sometimes our storage may be limited. For example, we may say that we do not have this 284 million cubic meter of available storage. The maximum we have available is only 200 million meter cube. So, the problem which we now have is, suppose we have the storage available to us 200 millimeter cube, what is the maximum demand which we can meet with this storage? So, for that purpose, we can do a number of things. We can do iteratively in the graphical method also. We have this S as 200. So, we can take this difference and draw tangential lines here and find out what is the rate which gives us S exactly equal to 200 or we can do it. These days, computers are easily available and there are lot of facilities which we can give any value here, find out this S and then we can say that make this value equal to 200 and find out what is this cube. So, in that case, we start with any value. Suppose, we have started with this 50 and we have obtained minus 284.1 or we have started with 45 and we have obtained this 205.1. One option would be to do some kind of interpolation between 45 and 50 and 45 we have 205.1 at 50, we have 284.1. So, little extrapolation will give us the value for 200, but using the spreadsheets, we can put any value here. This value will be different from 200 and then we ask the computer what should be this value to make this 200 and in this case, it turns out to be that this value is 44.5. So, this is the demand which can be satisfied for 200 million cubic meter of storage. So, using these techniques, we can determine either what is the storage required for a given demand or what is the demand which can be satisfied for a given storage. So, we have looked at the problem of finding out the storage for a supplying required rate of water or we can find out what is the maximum rate which we can supply for a given storage. Now, we move on to discussion of hydrographs which we have seen are record of the discharge in a river with time and suppose we have data given like this. This is time in hours and for a particular rain, the intensity is given here. We obtain the runoff as q. So, this is the total runoff which includes the base flow and the direct runoff. The intensity of rainfall is given here so we can see that at 4 hours. So, let us look at 0, 4, 8 and up to 12. So, from 0 to 4, we can see that there is no rain. Then the intensity is 8 millimeter per hour. Now, if we look at this at time t equal to 5, the intensity is 8, 6 it is 8 and 8 it is 8. Then at 10 hours and 12 hours the intensity is 6. So, by plotting at 4 it is 0, but if we look at the runoff, we see that the runoff starts increasing at t equal to 4. So, from 4 it goes up to 5 that means the rainfall has started roughly at about 4 hours. So, instead of showing it at 5, we say that from 4 to 8 we have a intensity of 8 millimeter per hour and then from 8 to 12 we can idealize this as intensity of 6 millimeter per hour. So, this is our rainfall, which is causing this runoff. Of course, rainfall will cause the direct runoff and base flow will come from the ground motor contribution. So, if we plot the total runoff, so this is the runoff it starts at 5, then it starts to go down if we look at these values 5, then 4.5 and 4. So, it goes down till the rainfall starts. So, this is the point at which the rainfall has started and then the runoff starts rising. So, if we look at the data we can see that at 12 hours it reaches a peak of 28 meter cube per second. Which is shown here and then again it starts to decrease because rain has stopped and it will continue till it becomes almost constant at about 4 after 64 hours. So, we get the runoff or the hydrograph, the runoff hydrograph as this typical curve. Now, our aim is to derive the direct runoff hydrograph, which means that we have to somehow separate the base flow component and we also have to find out the effective rain. So, this hydrograph which we have shown here, this is the total rainfall part of it will go as infiltration or evaporation. So, that loss as we have seen earlier is given by the phi index in a very simple average loss function. So, we will see how to obtain the phi index, how to separate the base flow and obtain the DRH. The first starting point of course is quite clear that wherever it starts rising we can say that this is the start of the DRH. As we have seen here the base flow contribution is decreasing at a certain rate. One of the methods of separating the base flow is just taking a constant value of base flow for example, this and saying that everything above this line is the direct runoff. But that method is not very accurate. So, we will be using a method in which from the peak we take a distance of n where n depends on the catchment area. In this case, let us say that the catchment area A is 30 square kilometers. The value of n based on various observations has been given as if A is in kilometer square then n comes out to be in days. So, for an area of 30 square kilometers we get n about 40 hours. So, that means the direct runoff will stop 40 hours after the peak. So, in this case we can get some point here which will be 40 hours from the peak. The method of base flow separation is that the ground water contribution is continued till the peak. So, just below the peak we will continue it and then we will join the point after n days directly with this point. So, using this method of base flow separation we can obtain the d r h by taking this q total runoff and subtracting the base flow which is done here. So, corresponding to total runoff 4 hour ordinate we get 0 because this is the start of the d r h. So, d r h starts here corresponds to t equal to 0. The ordinate of course is 0 then at 1 hour which is this point we take the ordinate which is 5 meter cube per second and subtract the base flow contribution which is computed from here and we get a d r h of 4.75 saying that the base flow contribution at this point is 0.25. So, these ordinates give us the direct runoff and we can see that the total discharge maximum discharge in this case is 26 which is 28 minus this 2. So, in this way we can obtain the d r h and plot it and this figure shows the d r h. This d r h is because of a rainfall which is given by 2 4 hour 2 4 hour rains of intensity 8 millimeter per hour and 6 millimeter per hour. Now, we have to find out the phi index and to do that we need to find out what is the volume of direct runoff and that computation can be done based on the d r h ordinates. So, here we have shown the ordinates of the d r h at t equal to 0 we have 0 then 1 4.75 at 2 8.5 and at 52 it stops here. So, for a 4 hour rain the runoff will continue for about 52 hours. Now, we can find out the volume which is accumulated at different times this is in 1000 meter cubes. So, at time equal to 0 then we have 0 volume at 1 hour we can say that this corresponds to the mean between 0 and 4.75 for 1 hour duration. So, if we do that we get 8.5 5000 meter cubes and here we have cumulative from the beginning of the rain. So, doing this for each time step you will notice here that the time steps are not uniform here we have 0 1 2 and then it becomes 4 6 8 10 12. So, we have a difference of 1 here then 2 and then it becomes a difference of 4 up to here last point we have we can interpolate 148 also here, but in this case it is 8 hours here. So, 8 hour difference then this 4 hours 2 and 1 generally the computations are much easier if we have equal spacing of the coordinates, but in this case since the rising limb is very steep we have decided to take small intervals on this and therefore it is not uniformly spaced computations are little more difficult in this case, but we can still do it using the spreadsheets. So, let us look at this volume for each time step and cumulative volume. So, for the entire direct runoff hydrograph the volume is given as 1.43 million cubic meters. So, this 1.43 million cubic meters on an area of course is given 30 and the volume. So, we want to find out what are the losses. So, we can find out what is the excess rain volume in this case the volume of the rain will be same as the volume of the DRH and therefore, we can convert into a depth would be 1.43 10 to power 6 divide by 30 kilometer square we can convert into meter square. So, so many meters and this turns out to be 47.7 millimeters. So, this means that the DRH which is shown in this figure this DRH is the result of 47.7 millimeter of rain over the entire catchment area of 30 square kilometer. The assumption of course is that the rainfall occurs uniformly over the entire area in that case it should have a depth of 47.7. If you look at the rainfall hydrograph the depth total depth is 8 millimeter per hour into 4 millimeter 4 hours plus 6 millimeter per hour into 4 hours equal to 56 millimeter. So, out of the total rain of 56 millimeters the effective rain or the excess rain is only 47.7 millimeters. So, that means we have a loss of 56 minus 47.7. So, 8.3 this 8.3 millimeter loss is occurring over a time period of 8 hours. Therefore, we can say that the phi index will be 8.3 divide by 8 which is 1.04 millimeter per hour. So, if our rainfall is like this 8 and 6 then we have a phi index of 1.04 that means our effective rainfall intensity for 4 hours this would be 8 minus 1.04 and this would be 6 minus 1.04. So, this is the effective rainfall 4 hour of intensity 6.96 another 4 hour of intensity 4.96. So, in this way we have obtained the phi index which tells us that effective rainfall intensity for the first 4 hour period is 6.96 or almost 7 millimeters per hour and for the second 4 hour period it is almost 5 millimeters per hour. Now, suppose we have a unit hydrograph available for the catchment. We can use that unit hydrograph to compute the direct runoff for storms which have a duration of 4 hours. So, for these 2 storms which are consecutive storms of 4 hour duration we can use a 4 hour unit hydrograph to predict what will be the direct runoff. So, let us assume that a 4 hour unit hydrograph is given to us. So, this is given a little later we will see how to obtain or how to derive this unit hydrograph, but let us for the time being assume that 4 hour unit hydrograph is given. So, you can look at the ordinates and this is the figure of 4 hour u h. So, as we have seen already 4 hour u h means this is due to 1 centimeter rain over a period of 4 hours which indicates that the intensity should be 1 by 4 centimeter per hour or we can say 2.5 millimeter. So, this hydrograph is a result of intensity 2.5 millimeter per hour continuing for 4 hours. So, if we draw the effective rainfall this is 4 hour duration and this is 0.25 centimeter per hour. So, this is 0.25 centimeter per hour or 2.5 millimeters per hour. So, because of this assumption of linearity we can now obtain the direct runoff for any rainfall of duration 4 hours or of a different intensity. For example, we have seen that the first rain 4 hour rain which we had has intensity of 6.96, the second had intensity of 4.96. So, if we want to use the 4 hour unit hydrograph to compute the d r h for these effective rain falls of 6.96 and 4.96 duration for 4 hours we can make a table like this where the first column shows first 2 columns show the 4 hour u h. The next column q 1 q 1 corresponds to a rain of intensity 4.96. 6.96 millimeter per hour while 4 hour u h we have already seen 4 hour unit hydrograph corresponds to 2.5 millimeter per hour. So, q 1 can be obtained by the 4 hour ordinate multiplied by this factor of 6.96 over 2.5. So, this 2.1 is nothing but 0.75 times 6.96 divide by 2.5 this will give us 2.1. Similarly, for 2 if you replace this by 2 we would get 5.5. So, in this way the first this column q 1 shows the result the direction of because of rainfall of 6.96 millimeter per hour for 4 hour duration. Now, the second rainfall is again 4 hour duration but the intensity now is 4.96. So, now not only we have to multiply with this factor 4.96 upon 2.5. So, q 2 has a multiplying factor of 4.96 over 2.5, but it is also shifted because the rainfall is starting after 4 hours. So, we have to shift the ordinate by 4 hours. So, whatever this 0 corresponds to this 0 at 6 hours the value which we get will correspond to 6 minus 4 which is 2 hours for the 4 hour unit hydrograph. So, 2 times 4.96 over 2.5 will give us 4.0. So, this 4.0 is the value at 6 hours. So, q 2 at 6 hours will be equal to 4 hour ordinate at 6 minus 4 hours into the factor 4.96 over 2.5. So, we can take any value we can take this t equal to 10. The value here 15 corresponds to 4 hour unit hydrograph at 10 minus 4 which is 6. So, 7.5 times almost factor of 2 which will give us 15. So, in this way we find out q 2 lag it by 4 hours and then we add these 2 to get the total direct run off because of this entire rain. Because of the principle of superposition we can add these 2 and get this value of q. So, in this way we can super impose any number of rains provided they are all 4 hour duration and we can plot this is the from 4 hour. So, this is the d r h the solid line shows the value this value which we have obtained from the 4 hour unit hydrograph and these symbols this we have earlier shown as the observed from the our analysis of measured data at the outlet. So, we can see that they are quite close, but they are lot of assumptions in the unit hydrograph theory and therefore, they are not exactly matching, but still it is fairly good match and therefore, a 4 hour u h can be used to obtain d r h for storms which are of 4 hour duration. Now, we can derive unit hydrograph for any other duration which is a multiple of 4 hours by this principle of superposition. For example, if we want to obtain 8 hour u h for 8 hour unit hydrograph what we can say is that the 8 hour u h can be divided to 2 4 hour segments 8 hour u h corresponds to a rainfall which has intensity of 1 by 8 centimeter per hour or i is 1.25 millimeter per hour which is exactly half of the intensity which we have used for 4 hour u h 4 hour u h has i equal to 2.5. So, we can add 2 4 hour u h. So, this is first 4 hour unit hydrograph shifted by 4 hours. So, if you look at these 2 columns 2.5 millimeter per hour 4 hour and 4 hour the this column represents the d r h because of a rain of 2.5 millimeter per hour for 4 hours this represents same thing, but shifted by 4 hours because rainfall is now starting after 4 hours. So, we have 4 hour unit hydrograph ordinate shifted by 4 hours. So, this value corresponds to 0 the value at 6 corresponds to 2 8 to 4 and so on. This shifting then allows us to sum them up and after summing we can get total q and divided by 2 because as we have seen the 4 hour u h has intensity of 2.5 8 hour u h has intensity of 1.25. So, after summing them up we get an effective rain like this and then we divide by 2 to get effective rainfall same as the 8 hour u h. So, in this way we can obtain an 8 hour unit hydrograph which plots this last column versus time. So, at different times what is the discharge which we can get for an 8 hour u h which means due to a rain of duration 8 hours and intensity of 1 by 8 centimeter per hour. So, getting 8 hour unit hydrograph or let us say 12 hour unit hydrograph if you want 12 hours then it is the same procedure because the 12 hour u h corresponds to a rainfall which is duration of 12 and intensity of 1 by 12 centimeter per hour. So, that the total volume is 1 centimeter. If we add 3 4 hour u h ordinate by shifting them by 4 hours we get effective rainfall 12 hour duration, but intensity of 1 by 4. So, after summing them up we have to divide it by 3 to get corresponding to 1 by 12. So, any multiple of 4 hours we can obtain the unit hydrograph directly from the 4 hour unit hydrograph, but for any other duration which is not a multiple of 4 hours let us say we want unit hydrograph for 6 hour duration or we want 3 hour duration or 2 hour duration. Then we cannot directly use this principle and we have to use what is known as the S curve. The S curve indicates rainfall of infinite duration. So, suppose we are talking about a 4 hour S curve the intensity would be 1 by 4 centimeter per hour, but the duration will be infinite. So, what it means is that we have a rainfall which is continuous with intensity of 1 by 4 centimeter per hour or in terms of millimeter 2.5 millimeter per hour. Now, we want to find out the direct runoff because of this rain the excess rain. So, we do again the same thing we assume that this is a lot of consecutive rain falls of 4 hour duration. So, if we have a lot of 4 hour consecutive rain falls we can use the shifting method again. So, the first column shows the 4 hour unit hydrograph we have omitted some data here and here. Then the second column shows the same data, but shifted by 4 hours. So, this is a shift 4 hour shift this is 8 hour shift 12 hour and so on. So, this is again the same data repeated, but shifted by 4 hours every time and then we sum them up. So, for example, this 21.9 is the sum of all these columns, 21.25 this is again the sum of all these columns and so on. So, in this way the last column here Q represents the effect of a continuous rainfall of infinite duration with intensity 2.5 millimeter per hour. As we can see in the end here the values will become almost constant and if we plot this S curve it would look like this. The constant value here corresponds to rainfall of intensity 1 by 4 centimeter per hour occurring over an area of 30 in this case we have taken the catchment area as 30 square kilometer. We can convert into meter cube per second and the value turns out to be about 21 meter cube per second. This means that if a rainfall of intensity 1 by 4 centimeter per hour occurs continuously over the entire area of 30 kilometer square at steady state it will give us a discharge of about 21 meter cube per second. Now, this shape if you look at this shape of the curve it is not a very smooth shape there are lot of wiggles here. This is because of our 4 hour unit hydrograph is not exactly correct we have some approximations there and therefore, in computations we can see these wiggles. So, if we want to make it smooth we would like to fit some kind of smooth curve like this and use that as our S curve. So, this S curve is because of rainfall of infinite duration now suppose we want the unit hydrograph for a 2 hour duration rainfall. In that case what we can do is we can shift this S curve by 2 hours which means that the other S curve is because of rainfall again of infinite duration, but shifted by 2 hour and then if we take the difference of those 2 d r h's the difference will be caused by a rain of 2 hour duration 1 by 4 centimeter per hour intensity. So, if we take 2 S curves shifted by this 2 hour duration take the difference of the ordinate that difference will be because of this effective rain and if we want 2 hour unit hydrograph then we will have to multiply it by 2 because this represents 1 by 4 into 2 centimeter equal to half centimeter of rain. So, that is what we have done here in the next table time this is the smoothed the smooth curve which we have fitted here the ordinates of that smooth curve are given here and then we lag it by 2 hours. So, this 0 corresponds to this 0 with the lag of 2 hours similarly this 50 at 10 corresponds to 50 at 8 and so on. This difference is multiplied by 2 so that we get unit hydrograph for 2 hour duration and this is plotted in this figure. So, you can see that at 2 hour unit hydrograph is obtained using the S curve. So, 2 different S curves 1 starting from 0 the other lagged and then we take the difference multiplied by 2 to get the 2 hour unit hydrograph. So, in this way we can obtain the unit hydrograph for any duration which is not a multiple of the given duration. It may be less than 4 hours it may be more than 4 hours for example, 5 hours also 6 hours also we can do the same thing we plot the S curve lag it for the given time period. Suppose, we want to derive a 6 hour unit hydrograph we can make the 4 hour S curve lag by 6 hours. The resultant difference is because of rain of intensity 1 by 4 over a period of 6 hours that means, total rainfall of 1.5 centimeters. So, whatever ordinate we get as the difference we have to divide them by 1.5 and we will get the 6 hour unit hydrograph. So, in today's lecture we have seen how to obtain the flow duration curve from the stage measurement at a given location. How to obtain reliability like what is the reliability of a discharge 85 percent or 90 percent. So, how much time it will be available in the river? In other words we can say that what is the 85 percent or 90 percent dependable discharge which will be equal or exceeded for a certain percentage of time. Then we have seen how to obtain a direct runoff hydrograph from the total runoff measured at a point by separating the base flow. There are lot of techniques of base flow separation but we looked at one particular technique which work very well in this case. After separating the direct runoff hydrograph we can find out the volume of direct runoff and therefore, we can find out the effective rainfall. And from the total rainfall and effective rainfall we can obtain a phi index which will then give us effective rainfall hydrograph or ERH. So, from an ERH and DRH we can obtain unit hydrograph which we will see little later. But suppose we know the unit hydrograph for a particular duration and we have rainfalls of the same duration occurring. We can obtain the total runoff direct runoff because of these rainfalls of same duration by using the unit hydrograph for the duration of 4 hours. So, we have seen how to obtain DRH given the unit hydrograph and we have also seen how to obtain unit hydrograph for different durations which may or may not be a multiple of the duration of the given unit hydrograph.