 In today's lecture we are going to look at some aspects of the unit hydrograph. We will see how to obtain unit hydrograph for a basin and how to use it to predict the runoff for any given rainfall. As we have seen the unit hydrograph describes the direct runoff due to a unit depth of rainfall uniformly distributed over the whole catchment area. So, let us look at a catchment area and we are measuring the runoff or the stream flow at point A. Now, suppose the rainfall occurs over the basin or the catchment the requirement is that it should be uniform all over the catchment and for the unit hydrograph the depth of rainfall should be unit which we take as 1 centimeter. The duration of the rainfall may be any given duration if it is a d hour rain then the intensity of the rain would be 1 over d centimeter per hour. So, as to get total rainfall depth of 1 centimeter. Now, if this effective rain when we say effective rain it means rainfall minus whatever abstractions have taken place. So, this is the part which directly contributes to direct runoff. So, we have already seen that the runoff consists of mainly two components base flow and direct runoff and the direct runoff is the part which occurs immediately or with a very little lag after the rainfall. So, what we are interested in is correlating the direct runoff d r h the direct runoff hydrograph with the effective rainfall hydrograph e r h. So, if a rainfall of duration d hour occurs with the intensity of 1 over d centimeter per hour indicating that the total rain effective rain is 1 centimeter and if we measure the direct runoff at point a that would give us the unit hydrograph of d hour duration. And it would look like this where the time may be in hours the stream flow is generally in meter cube per second. The use of the unit hydrograph is to estimate the runoff for any rainfall. For example, we can easily estimate what is the rainfall which is likely to occur once in 100 years and based on that pattern of the rainfall we can apply the unit hydrograph and estimate the runoff for that once in 100 year rainfall and we can use that to design flood control structures. So, let us look at the methodology of deriving the unit hydrograph for a basin. The first step is to select a duration d because the unit hydrograph changes with duration if we have a rainfall of longer duration the intensity would be smaller to maintain the unit area of 1 centimeter. If we have a rainfall of shorter duration the intensity would be higher in such a way that I 1 d 1 should be equal to 1 and I 2 d 2 should also be equal to 1 and naturally these two different rain falls will give us different hydrographs. If we have a shorter duration the time base will be smaller and the peak would be larger to maintain the same area. So, for duration d 1 we would have a longer base time base of the hydrograph and for the rainfall of duration d 2 we will have a smaller base, but a higher peak because the area under the hydrograph should be equal to the rain falling over the catchment. In fact, if we look at an ideal or theoretical unit hydrograph assume it to be rectangular then we can see that the area of the rain or let us say the volume of the rain falling over the catchment would be equal to volume of the direct runoff at point A. Now, let us just for example, take this Q peak flow as 100 meter cube per second and let us say the time base is 40 hours. Let us say that the rainfall which occurs is of 6 hour duration with intensity of 1 by 6 centimeter per hour. So, the area of the rain is 1 centimeter the hydrograph is 1 centimeter that is the effective depth of rainfall. This is falling over the whole catchment let us say that area of the catchment is A C and typically these areas are measured in kilometer square. Therefore, the volume of effective precipitation would be A in kilometer square. So, we will multiply it with 1000 to get in meter square into the depth of rainfall which is 1 centimeter. So, it would be 0.01 in meter. So, this is the volume of effective precipitation because we are saying that the 1 centimeter of depth falls uniformly over the entire catchment area. Now, this should be equal to the volume of direct runoff which we can obtain from this direct runoff hydrograph and if you look at this figure the area would be half into the peak flow. Now, this peak flow typically is in meter cube per second. So, we have to multiply it by 3600 to get in meter cube per hour and then multiply by the time base assuming this to be a triangle the area will be given by half into peak flow into the base and these should be equal and if we equate these two then the area of the catchment comes out to be 720 kilometer square. So, this tells us that the area under the direct runoff hydrograph and the volume of rain falling over the catchment should be equal. So, that gives us an idea of sketching the hydrograph in such a way that the area matches with the catchment area. To select the duration we have to define or we have to select what is the best or most likely duration of rainfall over the catchment. There are some empirical relations for example, one of the relations is that if you find out the basin lag which is defined as the time lag between the center of the precipitation this point and the centroid of the runoff. The time lag between these two points is known as the basin lag and typically a good duration for the unit hydrograph should be about one third to one fifth of the lag. So, after a study of various basins it has been found that this would be an ideal location and typically one fourth of the lag is taken as a good unit hydrograph duration d. Now, once we select the duration d the second step would be to select a few storms which occur over the catchment area with duration d. In nature typically the storms will not occur with exactly the same duration. So, we can say that if the duration of rainfall is within 10 percent of d. So, which indicates that it should be let us say 0.9 d to 1.1 d. So, we select a few storms in which we have uniform flow occurring over the entire catchment and the duration of rainfall is 0.9 d to 1.1 d. We may get three or four such storms and for each storm. So, we select the storms with duration within this 0.9 to 1.1 d for each storm we measure the runoff at. So, at the outlet of the catchment area which is this point due to a uniform rain over this entire catchment area falling in a duration 0.9 to 1.1 d. We measure the stream flow and once we get the stream flow we have already seen various methods of separating the base flow and get the direct runoff hydrograph. We have already discussed three different methods of separating the base flow. There is another empirical method which is sometimes used to locate the end of the direct runoff. So, if this is the total runoff the starting point a is easy to see because this is the point at which the hydrograph starts rising. The ending point d typically is estimated by i to see where the base flow has started and direct runoff has ended, but there are some empirical relations which give us the location of the point d and one of the commonly used equation is that from the peak the point d has a distance of n in days which is related to the catchment area where a is in kilometer square. So, if we know the catchment area which of course we know all the time we can estimate the distance between the end of direct runoff from the peak of the hydrograph n. For example, if area is about 100 kilometer square then n would come out to be about 2 days. So, if you have this hydrograph measured you can estimate the end of the direct runoff by using this equation separate the base flow by using any of the methods. For example, the straight line method and the rest of the curve would be the direct runoff hydrograph which would look like this. So, when we measure runoff at a we then estimate the direct runoff hydrograph using the techniques which we described just now. We also should be careful that the area under this curve which represents the effective rainfall should be close to 1 centimeter because that is what our aim is to derive hydrographs for 1 centimeter of rain. So, this should be 1 centimeter roughly, but even up to 4 centimeter the storms can be used and then once we find the area we divided by that ordinate. So, let us look at the next step. So, the number 4 step would be plot the direct runoff hydrographs and this suppose we have 3 or 4 different storms which we have selected and obtain the direct runoff hydrographs for those 3 under ideal conditions all the d r h should be same because we have assumed that the runoff is time invariant and linearly related. So, if we plot the d r h and when we say d r h these are all normalized means divided by the depth of rain or the area under the curve. So, all these d r h these are for unit rain for unit depth of rain ideally all of them should be same, but in nature we have a lot of variations all the assumptions which we have made in the unit hydrograph theory they will not be true. For example, the storm the storm may not be uniformly distributed over the area it may be moving it may not be covering the whole area at a time. So, this all the 4 3 or 4 whatever the storms we have selected they will have different d r h's and our aim now would be to average these and the best way to take the average typically is to take average of all the peaks. So, suppose we have 4 storms. So, we add all the peaks and divide by 4. So, this average peak will be marked on the curve and we also find out what is the time to peak average time to peak. So, similarly we find out a t p bar and this way we will locate the peak of the hydrograph. Now, we draw an average line looking at the individual hydrographs we try to draw an average line for the rising limb and then similarly an average for the falling limb. Now, once we average out all the variations in these 3 or 4 hydrographs this is an averaged unit hydrograph and the next step would be to find out the area under the curve which should be equal to 1 centimeter in for depth. Since, we have drawn it approximately the area will not be exactly 1 centimeter and then suppose the area is more than 1 centimeter we can reduce the area by moving our averaged curve a little bit. So, that the area becomes 1 if it is less than 1 then we would increase this and try to get a different curve. So, this will be an iterative process in which we have to draw a number of curves and fix the one which has an area of 1 centimeter of rainfall depth over the catchment area as the unit hydrograph. So, this way we can obtain the unit hydrograph for any basin provided we have storms of that duration they are uniform over the area and all the assumptions are satisfied. Now, once we get the unit hydrograph the next step would be to use the unit hydrograph to estimate runoff for any given rainfall. The simplest case would be when a storm occurs of the same duration as the one for which we have developed the unit hydrograph. So, suppose this is the E r h of intensity 1 by d and duration d and this is the unit hydrograph corresponding to duration d. So, we will call it d hour unit hydrograph. Now, to estimate d r h for any given how to estimate d r h for any given rainfall pattern for a given because the main purpose of developing the unit hydrograph is to estimate d r h for any given E r h. So, let us look at a few cases the simplest case which we are looking at now is a rainfall which occurs for the same duration, but with a different intensity. So, let us say case 1 would be d hour E r h with let us say intensity is i centimeter per hour and i is not equal to 1 over d because if i is equal to 1 over d then whatever u h we have derived would be giving us the d r h. So, now we have to invoke some of the principles which we have discussed earlier one was the principle of linearity. So, we said that if a rainfall of some intensity occurs and we have a d r h for that rainfall and then if a rainfall of let us say twice the intensity occurs then all the d r h ordinates will be doubled. So, the same principle we will use here we know that this ordinate of d r h since this is a u h this ordinate corresponds to i equal to intensity equal to 1 over d. Now, if we have a different intensity let us say i then this ordinate will be multiplied accordingly with that intensity. Generally before computers were so popular it was all done in tabular form. So, we use to make a table of time unit hydrograph ordinate generally generated by u and then the direct runoff hydrograph ordinate. So, at time t equal to 0 naturally u will be 0 then let us say at some time we can have the time interval same as d or we can have half d or 2 d whatever interval we want to take let us take the interval as d then at time t equal to d the ordinate of the unit hydrograph is u 1 at 2 d it is u 2 and so on. Now, if rainfall of intensity i occurs then the d r h ordinate will simply be u into i over d because 1 over d is the intensity for which we have the unit hydrograph and i is the actual intensity of rainfall. So, d r h ordinate will be 0 u 1 i over d u 2 i over d which is basically saying that this whole curve will shift up or down depending on whether i is more than 1 over d or less than 1 over d. So, if we look at a case where i is 1.5 d 1.5 over d the hydrograph will look like this all the ordinates will be multiplied by 1 and half. Now, this was the simplest case where the duration of the rainfall was same as the duration for which we have derived the unit hydrograph there may be cases when duration is more than d or less than d. The next case which we will discuss is suppose we have a rain our e r h is now of duration 2 d. So, we have a d hour unit hydrograph and we want to estimate the d r h for a rainfall of duration 2 d and intensity again let us assume that there is a constant intensity i to make the calculations easier. We would assume that this i is 1 over d, but this i can be anything here we would use again the concept of time invariance and linearity. So, if we use the concept of time invariance and linearity it leads us to method of superposition. So, we say that if there are 2 different storm events occurring then we can add the runoff due to each single event and get the total runoff. So, if this is the u h d hour u h for the first storm it would work fine, but for the second storm we can think of the second storm as another storm is starting at time t equal to d and then the direct runoff hydrograph for this would be identical, but shifted by duration d. So, all the ordinates of this curve would be shifted by d and then to get the total runoff we just add these 2 curves and get the total runoff. For example, in this case up to d the curve would look exactly the same as the u h because there is no second curve, but beyond that we need to add these 2 and get a different curve and the time base will also extend up to the end of the second u h. So, in this way again we can do it in a computer very easily, but earlier we used to do it using tabular forms. So, if we do the tabular form calculation in this case we have the u h is known to us. So, we know the ordinates of now I am showing here 0 d 2 d 3 d these do not have to be at d they can be at half d they can be 2 d intervals, but let us for simplicity assume that the ordinates are tabulated at every d. Now, if you look at the procedure here in the addition what we are doing is we are shifting the second curve by an amount d. So, we would call shift and write a shifted u for the first and second times it would be 0 because now we are shifting all the ordinates by an amount d and then for 2 d which is the time here the ordinates of the second u h would be u 1 which was the ordinates of the first curve at time d. So, all the ordinates are now shifted and then we have to add these 2 to get our d r h. So, on this was for the case when the intensity of rain was constant and was equal to 1 over d. If we have a different intensity then we can accordingly multiply the u h ordinates by the intensity. For example, let us say that the e r h looks like this where this is intensity i 1 this is intensity i 2 the duration is still we assume to be same. And a multiple of d of course, because all this terms we consider here or occurring over a period of d. So, that we can use the d hour unit hydrograph and this is the d hour unit hydrograph. So, in order to find out the runoff due to this combination of rainfall we would again use the same tabular form, but now we would multiply the ordinates by i 1 over d. So, in this case the table would look like this you have t u you can shift it and then the d r h would be the first u h ordinates multiplied by i 1 over d plus the second u h ordinates multiplied by i 2 over d. So, 0 u 1 i 1 over d u 2 i 1 over d u 2 i 1 over d and then plus u 1 i 2 over d and so on. So, in this way even if the intensities are varying we can find out the d r h in tabular form. And these days with advent of computers it is very easy to do these computations in the spreadsheet that we will not discuss here, but the procedure remains the same that you will shift the u h multiply by the intensity and add them to get the d r h. Now, these are simple cases when the duration of the rainfall was same as the duration of the unit hydrograph. There may be cases when the rainfall occurs for a different duration and in general the rainfall which occurs may not be the intensity of the rainfall may not be constant for a duration which is very long what we would do most of the times is assume that it remains constant for some time. So, for example, if the rainfall hydrograph is given like this we may assume it to be constant for some time. So, we may idealize it like this and then based on that we can analyze and find out the d r h. Now, if the rainfall which occurs if suppose we idealize this as d hours rainfall then we can use the d hour u h, but if the rainfall cannot be idealized like this suppose the rainfall looks like this then we may be able to idealize it like this in which the intensity of rainfall is a constant for a duration which is let us say smaller than d or not a multiple of d. So, suppose this is just for simplicity let us make it d by 2. So, now the question is how would we obtain the d r h for this rain because the d r h which is given to us has a duration of d and this rainfall which is given e r h has a duration which is smaller than d. Now, to solve this problem we introduce a concept which is known as S curve or S hydrograph. And what this hydrograph is is the direct runoff hydrograph for rainfall of a certain intensity occurring for infinite time. So, if the rainfall occurs for a very long time uniformly over the entire catchment then the resulting d r o d r h would be called S hydrograph and typically it looks like this and since this looks like an S that is why the name S hydrograph or S curve. The e r h for this case is a rainfall of infinite duration with certain intensity let us say 1 over d. So, if we obtain the S curve for a certain given rainfall let us look at how to obtain the S curve for a given duration d. If we know the d hour unit hydrograph so d hour unit hydrograph is given to us and the rain which is causing this is of duration d and intensity 1 over d. Now, suppose another rainfall occurs of the same duration and intensity 1 over d. So, if we have a d hour unit hydrograph available to us and we want to derive the S curve for this. We can say that the rainfall of duration d repeats itself a number of times actually till infinity. So, if we have a d hour and we have already seen that using the principle of superposition we can draw all these unit hydrographs shifted by the interval d and a large number of them. So, what it means is that if we add all these ordinates we would get the S curve. So, in the tabular form we can perform the S curve calculations in the same way as we have done earlier. For example, if we have the ordinates of unit hydrograph at every d hours we can shift it, but now we have to add a lot of these ordinates. So, shift by d 0 0 then u 1 u 2 u 3 and so on. Then we shifted by another d or shifted by 2 d and will have 0 0 0 u 1 u 2 and so on. And similarly a lot of them and then when we add them all up we would get here 0 u 1 u 1 plus u 2 and so on. And ultimately we would get this kind of this is known as the S curve and it represents the runoff direct runoff due to intensity 1 by d centimeter of the d hour. Runoff of rainfall occurring over the whole catchment for infinite time. Now, once we have this S curve we can derive runoff for any duration which is not a multiple of d, again if we look at the effective rainfall, suppose this is effective rainfall for a duration which is d by 2 the intensity is i. Once we know the S curve we know the S for duration d or intensity 1 over d. Let us assume that the S curve is given by this which represents infinite duration of E R H. Now, what we say that if we lag the E R H by this time d by 2. So, in the first case our E R H starts from 0 in the second case the E R H starts from d by 2. Then we would get S curve which would be shifted by the amount d by 2 and would be parallel to the first S curve. The difference between these two S curves at any time is because of the difference of rainfall between the first and the second case which is nothing but a rainfall of intensity d by sorry duration d by 2 and intensity. So, to summarize if we have a d hour S curve we shift it by d by 2 find out the difference in ordinates that difference will give us the hydrograph due to a rainfall of intensity 1 over d over a duration d by 2 and once we have this we can find out for any other intensity I what would be the ordinate. So, if we take the difference of these two S curves you will see that the difference is first increasing and then it is decreasing and becoming 0 at certain point. So, if we plot it would look like this and this direct runoff hydrograph is because of a rain of intensity 1 by 2 1 by d over duration d by 2. So, we multiply each ordinate by I over d and we will again get the required direct runoff hydrograph because of a rainfall of intensity I and duration d by 2. So, this again we can do it in a number of ways we can first derive the d by 2 hour unit hydrograph which is nothing but this is due to rain of d by 2 hour and intensity of the rain is 1 by d because this is the difference of the S curve which is for a d hour hydrograph. Now, d hour unit hydrograph has intensity of rainfall intensity of 1 by d centimeter per hour and therefore, this hydrograph represents a d by 2 hour hydrograph with intensity of 1 by d. Therefore, the volume of this will not be equal to 1 centimeter of depth it will be d by 2 into 1 by d which is half centimeter depth. So, this rain this direct runoff represents half centimeter of rain over the entire catchment. So, if you want to find out the unit hydrograph for d by 2 hours we will have to multiply all the ordinate by 2 and get a d by 2 hour unit hydrograph. So, in this way for a given d hour unit hydrograph we can derive u h for any other duration. If that duration is a multiple of d then it is very easy we do not have to go through S curve. So, let us look at deriving u h for any other duration. So, in a way we have already discussed this, but let us summarize this when we derive the u h for any other duration. If the duration is n d where n is some integer value then we just add up a lot of unit hydrograph suppose n is equal to 3 then we draw 3 unit hydrograph shifted by d add them up and get a direct runoff hydrograph. This would be d r o direct runoff ordinate would correspond to rainfall of duration n d in this case 3 d and intensity 1 by d. Therefore, the volume of rain which causes this direct runoff hydrograph will be n d into 1 by d equal to n centimeter and therefore, once we get this d r h we have to multiply by 1 by n or we divide each ordinate by n. So, in this case since n is 3 all the ordinate have to be divided by 3 and we would get a u h which corresponds to n d hours. The second case is if duration is equal to some fraction x into d where x is not an integer then we have to use s curve as we have discussed we can get the unit hydrograph for any other duration by drawing the s curve and shifting it by x d. Now, these direct runoff ordinate they are due to a storm of intensity 1 over d and period 1 over x d. Therefore, rain volume or depth of rainfall would be x and therefore, in order to compute the unit hydrograph for x d we will have to divide all these ordinate by x. Now, this is this works well when we have a rainfall hydrograph which can be idealized as rain falls of constant intensity over certain duration. We have also seen that for derivation of the u h we need rain falls of similar kind where we have a constant intensity d hour rainfall. In some cases it is not possible to get these very ideal kind of storms and sometimes it may happen that we have rain occurring over d hour periods, but with varying intensities. So, if we have the runoff corresponding to these complex storms we can still derive the unit hydrograph, but then the computations become a little bit complicated. So, let us look at a complex storm in which the e r h is given by all these are d hour duration, but the intensities are different. So, let us say that this e r h has depth d 1 this has d 2 this has depth of d 3 and what we have is we have the measurement of the runoff at the gaging station a which looks like now from this measurement of direct runoff hydrograph what we want to estimate is what will be the unit hydrograph for a d hour duration rainfall. And to do that we again use the principle of superposition. So, we say that let us assume that there is a unit hydrograph corresponding to duration d which looks like this and it has ordinate u 1 u 2 and so on which we do not know. Now, this is kind of inverse to what we were discussing till now and what we want now is to estimate the values of u 1 u 2 etcetera given the direct runoff values let us say q 1 q 2 at intervals of d. So, using the principle of superposition we can write equations like q 1 since the depth of rainfall for the first storm is small d 1 we can write the ordinate of d r h after time d as d 1 u 1 because the second storm has not yet started. Once the second storm starts we will have to account for the second storm also and now this is the first storm contribution because now it has already lasted for 2 d hours. So, the unit hydrograph ordinate at 2 d into the depth of the first storm plus the second storm has only lasted for d hours. So, it is ordinate u 1 multiplied by the depth of the second storm similarly we can write q 3 q 4 these are known to us from the direct runoff hydrograph which we have measured at a. So, now the problem is knowing q 1 d 1 q 2 d 2 and so on we have to estimate the values of u 1 u 2 and u 3 and so on. So, the procedure is very simple that first equation will give us the value of u 1 because q 1 and d 1 is known once we know u 1 in the second equation we can find out u 2 because then u 1 d 2 d 1 and q 2 all are known. So, the procedure is simple, but sometimes it leads to oscillations and the computed ordinate u 1 u 2 u 3 u 4 as we go towards the tail towards the end some of the use may become unrealistic some of them may become negative because we are going in a sequential order and the any errors in the computations will add up and will cause non physical behavior of the curve. So, sometimes we just fit even if we get negative values here we just fit and make it a smooth curve which has a unit area under the curve. So, to summarize what we have discussed today we have looked at unit hydrograph which is nothing, but the direct runoff produced by unit depth of rainfall occurring uniformly over the catchment area in a certain given duration. This duration is important for the hydrograph therefore, all the hydrograph we say we denote their duration of e r h. So, we say that it is a 6 hour unit hydrograph or 8 hour unit hydrograph or 4 hour unit hydrograph. We have looked at the ways to derive the unit hydrograph for a given catchment from a simple storm as well as from a complex storm. We have also seen once we have derived the u h how to use it to predict direct runoff for various other storms of different durations and in density.