 In the last lecture, we had looked at some of the limitations of the unit hydrograph theory and we have also looked at ways to go about it, how to solve these problems. For example, if the catchment area is very large, we cannot apply the US theory and then we can subdivide it into smaller areas, apply the unit hydrograph theory and route the flow through the channel or if the rainfall is non-uniform, we can derive an instantaneous unit hydrograph IUH and apply that to get the direct runoff because of any non-uniform distribution of the rainfall. In the third problem which we faced, there was a lack of data. So, if the stream is not gauged at a particular point or if storm events of that particular duration are not available, then we cannot develop the unit hydrograph for that catchment. For that purpose, we already described methods known as synthetic unit hydrographs which correlate some of the hydrograph properties to some of the basin properties and since the basin properties can be easily obtained, we can obtain the synthetic unit hydrograph for that area using some relationships which are mostly empirical based on observations of existing gauged catchments. So, let us look at a catchment where there is no gauging station, the point A is not gauged and therefore, we do not have any data about the runoff due to corresponding a storm event. In absence of this data, we can try to correlate some of the features of the hydrograph. So, let us say that the duration of the rainfall is t r, we can define a time to peak from the centroid of the rainfall as t p. Based on a lot of data about existing gauged catchments in the eastern United States, Snyder in 1938 came up with some relationships which relate the lag t p which is the time from the centroid of the rainfall to the peak with catchment properties. So, some basin properties can be used to obtain t p. What Snyder did was the basin properties he described in terms of two main variables length of the basin, this is from the point A, the outlet point to the water divide along the stream. So, this length is not a straight line distance between two points, it is along the stream a curving length l. The other parameter which he chose was based on the centroid of the area. So, if this point C is the centroid of the catchment area, then the point directly opposite the centroid on the stream let us call this point C prime. The distance of C prime from A again along this was termed as length to the centroid and we would write it as l c. So, using these two parameters l denotes the size of the basin because larger l value means larger size of basin and l c represents the shape of the basin. So, if the basin is wider near the point A l c would be smaller, if the basin is wider away from point A then l c would be larger and what Snyder found that the time to peak from the centre of the rainfall was related to l l c to the power 0.3. So, for any catchment if we know l and l c which we can measure from the topographic map of the area, then we can find out for a hydrograph of duration t r unit hydrograph of duration t r which was related with the time to peak as t p over 5.5. So, this Snyder called standard duration of rainfall t r and for this t r the time to peak will be given by this equation. Now, in general we may not want to derive unit hydrograph of duration t r and we may have the value of c t typically is obtained from a gaged catchment which is similar to the catchment under consideration. So, c t is obtained from nearby gaged catchment or gaged area. So, in order to derive the synthetic unit hydrograph we should have an area which is similar to the area under consideration may be a nearby basin where the properties are similar to the area under consideration and for that area we should have a gaging station from which we can derive a unit hydrograph for that particular area. So, let us look at a big basin and let us say that we have one stream here another stream here we know the catchment areas of these streams at some points let us call these points A and B and suppose there is a gaging station at point B. So, due to a rainfall event in the area B we can measure the runoff at point B, but at A we do not have a gaging station. So, our aim is to derive the u h at A and I will call it synthetic unit hydrograph S u h and this S u h may have a duration which is not equal to the standard duration t r it may have some other duration let us call it t capital R the required duration. So, this capital R stands for required duration now the data available at B again the rainfall event for this catchment area may not be of standard duration t r. So, let us say that at B we denote the properties by variables which have a star. So, the area of catchment B let us call it A star also let us say that we have a unit hydrograph derived for this area based since this is a gaged catchment area we can derive the unit hydrograph for this area based on storms occurring in that area and suppose we have derived the u h for a rainfall duration t star and the time to peak for this u h is t p star now this t r star is the duration of rainfall for this unit hydrograph which has a time to peak of t p star. Now, based on these data what we now want is to obtain the coefficient c t what Snyder said that if there is any other duration let us say t r star which is not equal to t r the standard duration then the time to peak will also have to be shifted. So, if for a standard duration there is some time to peak and if there is another rainfall which is not of a standard duration let us say this is longer than a standard duration then the peak will shift to the right and the distance to peak t p star. So, if this is t r and t p star will not be same as t p and there is a relationship which is proposed by Snyder that t p star should be taken as t p plus t r star minus t r over 4. So, this accounts for the difference in duration of the rainfall is standard duration t r and the actually occurring on the catchment area of b t r star. So, this helps us in obtaining the lag for the catchment area b now using these values we can derive the coefficient c t for the area b. So, what we do is we have this equation t p star and we also have the standard equation which says that t p is equal to 5.5 t r as proposed by Snyder. So, using these equations we can correlate the standard time to peak with the observed values of time to peak and rainfall duration as t p equal to 22 by 21 t p star which is rather straight forward operation by just substituting this value here and solving the resulting equation. So, using this information t p star duration t r star these are known because the catchment is gauged and therefore, we have unit hydrograph for duration t r star and we also have the time to peak in terms of t p star. So, we can find out t p and then since we know the relation of Snyder t p is equal to c t and since this is on the catchment area b I will put a star to the power 0.3. So, using this relationship we can obtain c t because once t p is known from here l star and l c star from the basin we already know this information because we have the data available for the gaged basin therefore, we can get c t from here. So, using this c t then we can obtain for we can assume that this same c t for b this is c t value for catchment b, but since we have assumed that both the catchments a and b are similar we can say that the same value of c t can be used for a also. So, may be used for a. So, that takes care of one of the coefficients c t now to find out the peak discharge Snyder found another relationship and he says that q p will be given by some other constant c p 2.78 a over t p. Now, this 2.78 factor comes by conversion of units of kilometer square and hours to get q p in meter cube per second. So, we take t p in hours a in kilometer square q p in meter cube per second and this c p again can be obtained from the data on the gaged catchment. So, let us we know the value of q p star we have the value of a star and we also have the value of t p. So, we can write c p and since everything else is known we can obtain c p for b and assume that the same c p holds good at a also and same as c p for a now this coefficient c p can also be written as if we look at discharge the peak discharge the peak discharge due to a rainfall of duration t r which just says that instances should be 1 over t r because we are talking about unit hydrograph and a p is the area which is contributing to the peak flow contributing to peak flow. If we want to write this in terms of Snyder's equation which is q p equal to c p what we get is the value of c p as t p by t r into a p by a t p by t r as per Snyder is equal to 5.5 a p by a will generally vary from catchment to catchment and therefore, the value of c p also varies depending on the value of a p over a and it has been found to vary from about 0.4 to 0.8. So, once we have the value of c p and c t. So, we have obtained these values from b and now we can use these values at a to derive a unit hydrograph for any desired duration t r. So, at a we want to derive a u h for duration t r capital R. So, now we have to see how to go about deriving the unit hydrograph at a for a duration t r. If this t r is equal to the standard duration t r then we do not have to do anything, but if it is not then again we have to use the Snyder's proposal and what we do is write t p r where this term is nothing but t p and then we can write q p for the required duration t r as. So, this will give us the time to pick for the required unit hydrograph and the discharge at peak q p and draw the unit hydrograph, but it will not be very accurate what we have is only two points this is q p t r and t p r. So, we have one point here, here we know it will start at 0 0 the other information which we have is that the area under the curve should be equal to 1 centimeter of depth over the cashment area. So, we may have a shape like this or like this. So, there are various other options also which all will have the same area. Therefore, some other points need to be given one of the parameters which is normally given is the time base t b this is the time base of the u h and there are again equations provided for this time base in order to help us sketch the unit hydrograph. There are some other parameters for example, the width of the unit hydrograph at some location can be given in this case suppose q p is the peak flow, we can find out what will be the width of the unit hydrograph at 50 percent of q p and call this w 50. Similarly, we can also provide some equation for w 75 which is at 75 percent of q p. So, once we know these 4 or 5 different parameters it would be much easier for us to sketch the unit hydrograph. So, let us first look at this t b the time base what we do is we assume that the unit hydrograph can be approximated by a triangle which has a peak flow of q p and a time base of once we assume it to be a triangle it is very easy to find out the area under the curve which knowing that q is in meter cube per second and t is in hours can be written as the volume can be written as half q p. Now, we have to take care of the units. So, this q p is in meter cube per second and t b is in hours. So, we have to multiply it by 3600 to get the volume in meter cube. Now, this should be equal to 1 centimeter of depth over the entire catchment area. So, this should be equal to area of the catchment into 1 and now to take care of the units a is in typically kilometer square. So, we have a million and this one is centimeters. So, it would be 0.01 in terms of meters and equating these two we can obtain an equation for t b in terms of the peak flow and the area and the equation is given as 5.56 into a over q p. Now, if the rainfall is not of a standard duration for required duration we will add subscript r here to show that this is the q p for the required duration of rainfall. Once we know t b then we have three points fixed one is the peak one is this 00 and one is this t b. For w 75 and w 50 there are some equations proposed again based on available data and the equations which are proposed are for w 50 2.14 over this is the peak discharge per unit catchment area. So, q p r over a to the power 1.08 and w 75 which would be smaller than this. So, now we have these five parameters known. So, we have a point here 0 point here t b we have q p r for the required duration of rainfall and we also know t p r. So, the peak point is known the base is known at 50 percent and 75 percent we also know what is the width of the unit hydrograph. There are some suggestions as to how much of this width should be before the peak and how much after the peak for example, some researchers have suggested one third before peak and two thirds after. So, if this is the peak line w 50 can be put one third here and two thirds here some of the other researchers have suggested 2 over 5 and 3 over 5. So, in the ratio of 2 is to 3 some have suggested 1 is to 2 some have suggested 2 is to 3. So, once we know that then we have these additional points two thirds here one third here. So, we have one point here one point here another point here another point here the base here. So, now we have 1 2 3 4 5 6 7 points which we can join by a smooth curve and obtain the synthetic unit hydrograph. The check will be that the area under this curve should be equal to 1 centimeter of rainfall and if it is not then we will have to adjust some coordinate may be this T b we will have to shift this side or this side. So, that little bit of adjustment has to be made in order to maintain the area under the curve equal to 1 centimeter of depth. So, in this way for any catchment which is not gauged, but a nearby catchment has a gauging station and which is similar to the basin under consideration. We can develop a hydrograph T R our unit hydrograph based on the synthetic unit graph. There are some modifications to this synthetic hydrograph which have been proposed. For example, if we take this equation the CT parameter depends on the slope of the basin. So, some of the investigators have put the slope of the basin also inside this variable and then they have suggested using some formula which involves the variable L L C over square root of s and to the power n and some factor CT prime. So, these kind of suggestions have been there in the literature available to us, but we will not discuss those advanced techniques right now this description of synthetic unit hydrograph should be sufficient for this purpose. Now, synthetic unit hydrograph or instantaneous unit hydrograph or a regular unit hydrograph these all give us how the discharge varies with time. Based on these U H we can derive the D R H for example, if we know the intensity of rain is 1 and half the depth of rain is 1 and half centimeters. We can derive a D R H which will have everywhere ordinates equal to 1 and half times the U H ordinate and once we get the D R H we can add the base flow and get the total runoff hydrograph which would look like our aim of developing this total runoff hydrograph is to be able to answer questions like what is the maximum flow in the stream, what is the minimum flow in the stream, what will be the total amount of water which passes the stream for a fixed period of time or there could be lot of other questions like one question which typically people ask is what is the reliability of certain discharge. For example, let us say 85 percent of the time we say what is the discharge which we can assure will be available to us 85 percent of the time. So, to answer all these questions we would look at alternative ways of presenting the same information. So, let us choose a time period it may be 2 or 3 days, it may be a week, it may be a month, but let us say that the hydrograph during that time period is given by this total runoff hydrograph. Now, if there are some questions which are very easy to answer for example, if somebody ask what is the minimum flow in the stream, we can say that this is the minimum flow. Then if you are designing for a blood control structure you may have to answer a question what is the maximum flow in the stream that is also very easy to answer q max. Now, somebody might ask what is the average flow over this time period let us call it t naught. So, over this time period of t naught, what is the average flow in the stream or what is the total volume of water which has passed the stream at that location a in this time t naught. So, for this purpose there is a curve which we can generate from this hydrograph which is known as the mass curve and mass curve shows the volume of water cumulative volume of water passing through that point in a given time. So, we have t and let us say we are concerned about time period from 0 to t naught only. So, if you look at the accumulated volume let us say in meter cube time may be in hours, it may be in days or months, but let us for simplicity use a single storm hydrograph and see how we can develop these curves. So, at time t equal to 0 the cumulative volume will be 0 and then as water starts coming in the river the cumulative volume will be increasing. It will be increasing at a fast rate till the peak and then the rate of increase will become slower as the discharge decreases. So, we get a curve which is concave then convex then again it may become concave here which tells us what is the average rate of flow by taking the slope of this line. So, the slope of this line this will give us total volume coming in time t 0 and this line will tell us the slope of this line will tell us what is the average flow rate in the stream for period 0 to t naught. Now, this flow rate in the stream of course, we cannot satisfy this flow rate suppose we demand this flow rate we will not be able to satisfy it because some of the times our discharge is less than let us show the average flow rate somewhere here and call it q average. q average is nothing but the total volume v divide by t naught this is the average flow rate if we consider the entire period 0 to t 0, but as you can see here in this portion the supply in the river is less than q average the rate at which water is coming in is less than q average in this portion it is more and then again it would be smaller than q average. So, we will not be able to supply water at this rate at the outlet point if somebody wants to guarantee certain rate of supply that would be q minimum we cannot supply anything more than that because if you supply little more than that during some period of time you will not be able to satisfy the demand the way to supply this demand is to create a storage facility. So, we will look at that in little while the other question which we were asking is about how much percentage of time we would be able to satisfy certain requirement. For example, we may say what is the flow rate what is the q which can be guaranteed let us say for 50 percent of time. So, if we look at what is the q which can be guaranteed at 50 percent of the time means for a period of t naught by 2 the q should be more than that guaranteed q and for that we must draw a line somewhere here which will have a width of t naught by 2. So, in this t naught by 2 time our flow will always be more than q let us call it q 50. So, one question which is typically to be answered by a water source engineer is what is let us say q 50 or q 85 what is the flow rate which we can satisfy 85 percent of the time. So, that also we will look at by curve which is known as a flow duration curve. So, let us see that from a given hydrograph how can we obtain these quantities which are important for a water source engineer to know. So, we have a minimum flow q min maximum flow q max and let us say we are analyzing this period 0 to 0 and we want to say what is q 50 means what is the flow which can be guaranteed at least 50 percent of the time. So, this is flow which is and we use the term equalled or exceeded. So, at least this amount of flow q 50 would be available to us 50 percent of the time. So, in order to find out this q 50 what we will do is we will take a line which is of width t 0 by 2 50 percent and then we will have to shift this line up and down till this line fits exactly between the two curves and this will give us q 50. Similarly, if I want to find out q then we must see what is the flow which is equalled or exceeded 85 percent of the time. So, that means only 15 percent of the time the flow would be below that. So, if we let us draw this line here again. So, this will be 0.85 t 0 and again we fit that line which will be 0.85 t 0. So, this flow would be equalled or exceeded 85 percent of the time. So, this is an important parameter sometimes when we are creating facility for power generation typically we take this 85 percent dependable discharge as our design discharge. So, that means at least 85 percent of the time that much flow will be available to us. To answer these questions we use what is known as a flow duration curve which plots this percent against q and this percent of time when we write here this is percent of time equalled or exceeded and the curve would look like this. So, there is some minimum flow which will be equalled or exceeded 100 percent of the time q min there will be some maximum flow which will not be exceeded. So, it will have a 0 probability of 0 percent of the time equalled or exceeded and then from this curve we can answer all the questions like what is 50 percent q 50 what is q 85 and so on. Now, the flow duration curve and the mass curve they are both very critical in water sources or let us say power generation design. We would first look at the mass curve to find out how much storage we would require in order to satisfy a given demand. So, if we draw the mass curve which as we have seen would look like this during periods of high discharge the slope will be larger then as the discharge reduces the slope becomes flatter and then if another storm comes it will go up. Now, we are discussing this on a very short time basis may be a few weeks or a few days, but generally this analysis is done for the annual flows, but to understand the concept let us look at this curve first which let us say is in response to a storm with the d r 8 with the runoff given by this. So, this runoff will produce this mass curve where v is the cumulative volume as we have seen if this is the time t 0 there is an average flow which can be thought to occur during this time t naught and we cannot supply this average flow as we have discussed because some of the time the flow in the river is not sufficient and it will not be able to meet the demand. If we want to meet this demand suppose at point a we can create some facility for storing water typically it is a dam built on the river to store water, but we can assume that there is some structure here which stores water. Now, during periods of low flows when the flow is less than the demand we can get water from the storage to supply the demand or to supply the requirement which exists suppose at this continuous rate of course, we cannot supply water more than this rate our q supply this is the maximum rate at which we can supply water because this is the total amount of water available to us v is the total amount of water available to us in time t 0 we cannot supply more than this water. So, this is our maximum rate of supply less than that of course, we can always supply. So, we can always supply anything at a smaller slope, but more than that we will not unless we take water from somewhere else. So, let us say that we want to supply this maximum rate by creating some storage facility here. The question which is to be answered is what should be the capacity of that storage structure. So, from this figure we can see if we draw the mass curve again. So, the idea is that during period of low flows we will take water from the storage and during period of high flows we will divert the additional water from the stream to the storage. So, if you look at this figure of the runoff during this time our supply is less than the demand and therefore, we must take water from the storage during this period water will be going from the stream to the storage after satisfying the demand and again during this period up to t naught water will be taken from the storage to satisfy the demand. In order to find out this capacity this is the average flow rate what we do is we draw lines parallel to this and then the difference between these two lines is the required storage. These lines are drawn at the ridge and the valley. So, the highest portion and the lowest portion of the mass curve which is the volume versus time the difference will be the required storage. So, this S is the required storage to satisfy a continuous demand at a constant rate which is equal to the average flow rate. This is the maximum sustainable demand at that location and if you look at this figure it is clear that when the mass curve is flatter that means we are getting water at a smaller rate and therefore, water has to come from the storage into the stream to satisfy the demand. There are various other options or various other variations in demand for example, the demand may not be constant. The demand may not be as high as q average the demand may be smaller in that case our storage requirement will be smaller. For example, if the demand curve which in this case we have taken parallel to this q average line if it is flatter then we will have to draw a line which is parallel to this demand line and this will result in a smaller storage for that satisfying that particular demand. Now, the demand itself may be varying. So, your mass curve may look like this and the mass curve may look like this and the demand itself may look like this. So, instead of a continuous constant demand we say that the demand may be varying with time. If you are looking at a large term let us say annual time period let us say this is 1 year which is what we typically do in water sources project because the rain runoff all these things are cyclic based on an annual cycle. So, typically we look at a 1 year period and see sometimes the demand may be very high may be in summer months the demand may be high depending on the locality whether the demand is for irrigation or residential requirements. The demand may be high during summer it may be low during winter and so on. So, the demand itself may be changing water supply of course, changes. So, in that case what we do is we find out the maximum difference between the demand and the available water which is the runoff cumulative runoff volume. So, the maximum difference between these two ordinates will give us the required storage capacity. So, we will now discuss one more aspect which is related to the mass curve and typically we will talk about annual flow in the stream. As we have said looking at short term week or days is also possible, but generally when we say yield it would be annual yield of the stream. So, if you look at the hydrograph of the stream typically this time will now be January, February, March and so on. The hydrograph may look like this small discharge here then in monsoon months it will go up and then again in winter months it may come down. There may be a smaller peaks here where these in response to individual storms, but in general the hydrograph will look like this and yield would mean what is the amount of water available in the stream over the period of a year. So, this would be cumulative volume typically over the year. So, yield is an important concept because it tells us how much water is available to us over the period of the year. Hydrograph will give us what is the maximum flood in the river. So, we can use that to design flood control works. We also have this minimum flow here and as we have discussed earlier there are three different kinds of streams we have perennial and if you draw this hydrograph for perennial you will see that throughout the year there will be some water available in the stream. Therefore, if you draw the flow duration curve the 100 percent will correspond to some finite non-zero value of q, but if you have an intermittent or ephemeral stream then the flow duration curve would look like this where 100 percent may be here and this may be 70 or 80 percent. So, this means that 20 percent of the time there will be no flow in the stream. So, this is also an important aspect which has to be taken in consideration when we are designing a project that 20 percent of the time there will be no flow in the stream. So, if we are generating power there will be no power available during this time. So, we will finish here and in summary in today's lecture what we have discussed is how to develop hydrograph for engaged basins using synthetic unit hydrographs, how to develop mass curves and flow duration curves which help us in analyzing the flow in a river in terms of what is the dependability of various discharges, what is the storage required in order to satisfy some demand and what is the maximum demand which we can meet based on the given flow pattern.