 So, today we are going to continue with our description of the unit hydrograph. We have already seen how to derive a unit hydrograph, how to apply it to predict the runoff due to any given storm. Today we will see what are some of the limitations of this approach and how to go about getting other data to take care of some of these limitations. Let us look at some of the limitations. The first limitation is that the unit hydrograph approach is not valid for very large areas and when we say large areas typically areas more than 5000 kilometer square unit hydrograph approach will not be valid because lot of assumptions which we make in the UH analysis for example, uniform rainfall over the entire area they will not typically be valid for a very large area. So, typically we will say that upper limit of applying the unit hydrograph would be about 5000 kilometer square. The second limitation is that we have assumed that the rainfall occurs at a uniform intensity for certain duration. In practice an actual rainfall event may not follow a uniform intensity for a large large period of time. The actual rainfall pattern may look like this. So, even though we have some storm of let us say d hour duration the intensity will not be a constant and typically we may not be able to assume it to be a constant for certain duration. In that case also the unit hydrograph approach will not work very well. So, the rainfall non uniform and also it cannot be approximated also by a uniform distribution in that case we cannot apply the unit hydrograph approach. The third limitation is that we must have data for deriving the unit hydrograph. So, either if we have a catchment area and we want to derive unit hydrograph at point a we should have a gaging station at point a. If we do not have a gaging station at point a we will not know what is the runoff at point a or sometimes even if we have gaging station at point a we may not have required a storm occurring in that area. So, lack of data would be another limitation which will prevent us from deriving a unit hydrograph. So, today we will look at some ways of getting over these limitations or how to do our analysis if let us say some of these limitations are not satisfied. For example, the area may be very large rainfall may be very non uniform. So, how to go about it? We will start with the non uniform rainfall and how to analyze it. We use a method which is known as IUH or instantaneous unit hydrograph. I stands for instantaneous and UH of course, is the unit hydrograph. So, let us look at what is an instantaneous unit hydrograph and how do we use it to generate direct runoff for a non uniform rainfall. By now we are familiar with the unit hydrograph shape. So, let us say that we have a unit hydrograph for an EH of certain duration d and of course, the intensity would be 1 by d centimeter per hour. Now, if we reduce the duration of the rainfall and increase the intensity in such a way that the total amount of rain falling is still 1 centimeter we would get a different unit hydrograph. So, this is a 1 hour sorry a d hour UH. Now, let us reduce the duration to d by 2 and increase the intensity to 2 over d. So, in this case again we have a 1 centimeter of rainfall, but for a smaller duration and with a larger intensity. The unit hydrograph corresponding to this would have a smaller time base because the duration of rainfall has reduced, but will have a higher peak because the area of the hydrograph must remain the same. So, this would be a d by 2 hour UH. Same argument can be extended further if we further reduce the duration by half and increase the intensity to twice. We would get an ERH which is very narrow and the intensity is very large. In that case the UH ordinate peak will again be larger time base will become smaller. So, we may get a curve which has a smaller time base, but a higher peak. Carrying this argument further we can reduce the duration to 0 and that is why this is called an instantaneous unit hydrograph. So, this unit hydrograph is the result of a rain which occurs instantaneously. The intensity is infinite of course, the duration is 0. Naturally we will not be able to show it on this figure, but the intensity and duration are such that the total amount of rain is 1 centimeter and the DRH resulting from this is known as the IUH. So, as you can see that there is no duration attached with IUH. In unit hydrographs we had a d hour unit hydrograph for example, it may be a 6 hour unit hydrograph 4 hour or 2 hours, but IUH is for instantaneous precipitation with total depth of rainfall equal to 1 centimeter or sometimes people take 1 inch. So, our aim is to derive this IUH and then use it to analyze the direct runoff due to a non-uniform storm. So, let us first see how we can obtain the direct runoff due to a non-uniform intensity distribution. Let us say this is the ERH and the ERH may look like this. The total duration of rain let us say is T 0 and suppose we know the IUH for the catchment and we want to find out what will be the direct runoff at point A due to this ERH over the entire catchment area. Let us say that the IUH looks like this. Typically the ordinates of IUH are denoted by U. So, the small u t it represents the ordinates of the IUH at time t. Now, in order to analyze this ERH and obtain the DRH. So, our aim is to obtain a DRH which may look like this. So, this DRH is caused by this ERH subjected to this IUH and we will have to develop an equation to relate this DRH with the ERH and IUH. So, let us do it now and take a small ERH element. Let us say this is the ERH showing the time versus intensity of the rainfall. Now, this we can write as I tau where tau is the time from the starting of the rainfall. So, at any time tau the intensity of rainfall is given by I tau. Now, if we draw the IUH is starting from this point tau and want to find out the DRH at any particular time t. So, our aim is to find out the DRH ordinates at the point t which we will call q t. Now, this q t is because of this rainfall intensity and this IUH and if we draw the IUH starting from this point as origin that means the direct runoff due to this strip of rain can be given as the depth of rainfall here which is I d tau multiplied by the ordinate of the IUH at that point. So, this ordinate is nothing but u t minus tau. So, if you look at this dq, dq is the direct runoff due to this small strip of rainfall occurring over the entire cachement. So, and we say that this is the depth of rainfall I tau d tau occurring over the cachement instantaneously it will cause a direct runoff of dq and if we want to find out the total ordinate q t we have to integrate it from 0 to t. So, this equation gives us a way of finding the direct runoff ordinate at any time t. Now, if you look at this expression I tau will of course be 0 beyond t 0. So, there is the duration of rainfall is t 0. So, if t is more than t 0 then we have to perform this integral only from 0 to t 0 and this way if you look at another strip here. So, what we are doing by this integral is taking all these small strips and finding out their contribution to the direct runoff at the time t. So, every time we have to take a strip I tau d tau will be the depth of rainfall and u t minus tau would tell us what would be the contribution at time t due to that rain occurring at time tau. So, t minus tau will be the ordinate which we have to consider for the I u h. Now, if we should look at how to derive the I u h as we have already discussed it is a very small duration rain total depth of 1 centimeter. So, one way of deriving the I u h is using the S curve. Now, we have seen in the previous lecture this S curve is the direct runoff hydrograph because of an infinite duration of rainfall. So, the E r h looks like this and continues to infinity the intensity would be 1 by d centimeter per hour if we have taken a d hour unit hydrograph and as we have seen earlier what we do is we take a number of d hour hydrographs and shift it by d hours and this is a d hour u h. So, if we take a number of these and shift them at them we would get what is known as the S curve and let us it will reach a maximum ordinate here q max which can be obtained by the rainfall intensity and the catchment area. So, if you have a catchment area a in kilometer square and rainfall intensity of 1 by d centimeter per hour then q max can be given by a very simple expression in meter cube per second. So, here we have taken care of the units. So, since this is per hour we have a factor of 3600 since this is kilometer square we have a factor of 100 1000 1000 and since we have d in centimeter we have another factor of 0.01. So, if you put all these factors in here you will get the maximum ordinate of the S curve as 2.78 a over d. Now, once we get the S curve suppose we now shift this S curve by some amount let us call it delta t. So, this is an S curve obtained from a d hour u h. So, let us call it a d hour S curve and then we shift it by an amount delta t draw the e r h for the first S curve again e r h will go on to infinity and then the e r h for the second curve will start from delta t will have the same intensity 1 over d, but a difference of delta t in the starting point. So, if we take the difference delta s this delta s indicates the direct runoff due to a rainfall which is the difference of these two e r h which is nothing but an e r h of duration delta t and intensity 1 over d. Now, in the limit that delta t tends to 0 we will get the i u h, but we have to maintain 1 centimeter of rainfall depth. In this case the rainfall depth is delta t over d centimeters because the intensity is 1 over d and the time duration is delta t. So, if we have delta s due to a delta t over d centimeter rain we can find out what will be the ordinate due to 1 centimeter of rain by using the principle of linearity. So, the ordinate d r o direct runoff ordinate due to 1 centimeter of rain can be obtained from delta s delta t and d. So, let us write it in terms of first finite values d delta s over delta t and then we will take the limit as delta t tends to 0 we can get d s over d t. d s over d t is nothing but the slope of the s curve at a particular time and d as we have seen is the duration of the rainfall for the unit hydrographs. If we keep d equal to 1 then what we get is the ordinate of the instantaneous unit hydrograph at any time t will be d s by d t. So, we have a very simple method of deriving the instantaneous unit hydrograph ordinate u t by drawing an s curve which is for a duration of 1 hour rainfall. So, if we have developed a 1 hour u h we can derive the 1 hour s curve and then we can find out the slope of that s curve at different times and that slope will be equal to the ordinate of the instantaneous unit hydrograph. So, this is one of the ways of finding out or deriving the unit hydrograph there is another method which is commonly known as the Nash method because he was the first one to propose this concept and derive the i u h. In Nash method which is also known as a linear reservoir model he conceptualized the basin as a series of reservoirs such that the outflow from one reservoir goes to the next reservoir as inflow like this there are n number of reservoirs and if we put the inflow as 1 centimeter instantaneously whatever outflow we get from the n th reservoir would indicate the i u h. So, this is the basic concept of the Nash model that the catchment can be thought of as a series of reservoirs which are linear. Now, linear indicates that storage and discharge s and q they have a linear relationship in the reservoir typically if you look at a reservoir which has an outlet here the storage will depend on what is the height in the reservoir and the outflow will also depend on the height in the reservoir. So, both the storage and the outflow they are functions of the height in the reservoir and therefore, the concept of linear relationship between s and q has been used by a number of investigators they say that q and s can be linearly rated if you look at the dimensions of q this is volume over time meter cube per second typically and s is in terms of volume which is typically meter cube. So, if we use a time constant. So, q into k equal to s where this k is the storage time constant and it has units of time this k would be a property of the reservoir or in this case the basin. So, in Nash model we say that q and s are linearly rated s can be written as q into k where k is the time constant and then we can write the continuity equation for any reservoir if there is some inflow i and some outflow q we can write the continuity equation as the net inflow equal to change of storage. So, let us take a particular reservoir and consider a time period of delta t the inflow in time period delta t will be i into delta t and the outflow will be q delta t. Now, s is the storage within the reservoir and suppose in time delta t s changes by an amount delta s then we can say that this change in storage is because of the net inflow and it should be equal to the net inflow. So, we can write and then if we take the limit as delta t tends to 0 we get the differential equation of continuity what is commonly known as the continuity equation or the mass balance equation which tells us that d s by d t is equal to i minus q. Now, our aim is for any given input we want to find out what is the output q therefore, assuming the linear reservoir we can write this s in terms of k and q and therefore, we will get an equation describing the variation of q with variation in i and this differential equation can be solved this is a linear first order equation therefore, it can be easily solved to get the value of q in terms of i. Once we get the outflow from one reservoir that outflow let us call it q 1 becomes the inflow for the second reservoir. So, i 2 is equivalent to q 1 similarly, for other cases as once we get q 2 it will become the inflow for the second third reservoir. So, i 3 will be equivalent to q 2 and so on. So, ultimately our aim is to get the outflow from the n th reservoir q n and following this equation for solving this equation for each reservoir we can get q n which will give us the i u h ordinate because what we are saying here is that 1 centimeter of rain falls instantaneously on the first reservoir and then is carried through to the n th reservoir this q n represents the outflow at point a and therefore, since this is the outflow because of 1 centimeter of rain falling instantaneously the outflow will be giving us the ordinate of the i u h. So, this is the basic theory of the Nash conceptual model and if we solve these differential equations the solution of this differential equation can be written easily in terms of i and k. So, let us look at this solution of this equation for the first reservoir in which i is an instantaneous the inflow is instantaneous at time t equal to 0 we have a rain which is 1 centimeter intensity of course, is infinite duration is 0, but the total volume is 1 centimeter. So, this i for the first reservoir would be a delta function which occurs at t equal to 0 and therefore, the integral of this term would be nothing but the value of exponential t over k at t equal to 0 using the property of delta function the integral gives the function value at that particular point and therefore, this will be equal to 1. So, the outflow from the first reservoir q 1 is quite straight forward as t over k exponential minus t over k and then we can apply this outflow as the inflow to the second reservoir and now we want to find out what is q 2 equation is of course, the same the solution we have already described is 1 over k e minus t k integral of i exponential t over k d t, but this i now for the second reservoir will be replaced by t over k e minus t over k. So, if we perform this integration we get a value of q which is given by q 2 equal to 1 over k square t e minus t over k and as we keep on doing this for example, q 3 is and so on till we reach q n which comes out to be factorial n minus 1 k to the power n t to the power n minus 1. So, the outflow from the n th reservoir is given by this equation q n and since we say that this outflow is nothing but the i u h ordinate we can write this as u. So, the ordinate of i u h at any time t will be given by this where k and n now are some property of the catchment. So, number of reservoirs how many are there what is the time constant both of these will depend on the catchment area and there are various methods of estimating these k and n we will discuss that little later, but the idea is that this ordinate depends on only two factors k and n. If we look at the values of let us say n it does not have to be an integer although the way we have conceptualize n is the number of reservoirs, but it does not have to be an integer your conceptual model may say that you have two and half reservoirs in the base n or 3.5 reservoirs in the base n. So, in order to generalize this equation for non integer values of n we use the gamma function and the definition of the gamma function probably you already know this n minus 1 factorial is nothing but gamma n. So, in order to make it valid for non integer n values we replace factorial n minus 1 by gamma n and say that this would be valid now for any non integer value of n also. There are relations between k and n and the moments of the i u h. So, if this is the hydrograph then if we take the moment of this curve about the origin those moments can be shown to be functions of n and k. For example, if we take the first moment it is equal to n into k and if we take the second moment it is equal to n n plus 1 k square. So, if we have first and second moments of the i u h we can write these two equations in terms of n and k and find out the values of n and k for a given base n. Now, we will not have the i u h in general available to us because our aim is to derive the i u h, but what we may have is some e r h and corresponding d r h. So, if we have these e r h and corresponding d r h values we may find out the moments of these e r h and d r h and those can be correlated with the moments m 1 and m 2 of the i u h and the formula which are given for this case use four values m q 1 and m q 2. These are the first and second moments of the d r h about t equal to 0 and then divided by the area under the d r h which would be the volume of d r h. So, we say it should be divided by the total direct runoff. So, we normalize the d r h and take the first and the second moments we will call them m q 1 m q 2. Similarly, for the e r h we define two moments m i 1 and m i 2 these are the first and the second moments of e r h about t equal to 0 and again divided by total effective runoff. So, once we get these four moments m q 1 m q 2 m i 1 m i 2 we can obtain m 1 and m 2 which are the first and second moments of the i u h and these we can correlate with n k n n plus 1 k square. So, we have two equations and two unknowns which we can solve to get the values of n n k. The relations between these are given as m q 1 minus m i 1 which is the difference in the first moment of the d r h and e r h. This is equal to the first moment of the i u h and therefore, is equal to n k. The second equation the second moment difference m q 2 minus m i 2 can be correlated with the second moment of i u h which is n n plus 1 k square plus the e r h first moment. So, if we have known values for these four moments and this m i 1 of course, will be used here also. So, if we have these four values known from n e r h and corresponding d r h we can estimate the values of n n k and once we have n n k then we know that i u h ordinate is given by the gamma function exponential and all these. So, these two n n k are the parameters which will depend on the cache moment and may be obtained from an available e r h and d r h value. So, the i u h which we have discussed just now takes care of any non-uniformity in the rainfall where we cannot approximate it by constant intensity over certain duration. Now, second problem which we will look at is if we have an area which is quite large now there may be a river which is joined by number of tributaries and this area being very large let us say more than 5000 kilometer square. We will not be able to use the unit hydrograph to predict the direct runoff at A due to any storm occurring over this area because typically for a larger area the assumptions made in the unit hydrograph theory will not be valid. So, if we look at this area we can subdivide this into a smaller areas and then apply the unit hydrograph theory over the smaller areas for example, if the cache moment area of this tributary is like this then this area being a small we may be able to apply the unit hydrograph theory for this particular tributaries find out the flow at this point let us call it point B. So, once we apply the u h theory over this shaded area we can get the outflow at point B, but now this outflow at point B will go into the channel till it meets the outflow from another tributary let us say at point C. So, the direct runoff from the tributary will come at point B it will travel through the stream and reach at point C where it will again meet the runoff from this tributary. So, if we find out at this point B the hydrograph of course, the d r h at B if we find out direct runoff hydrograph at point B we will then have to add the base flow to it to get the total runoff. So, these two steps can be done using the unit hydrograph theory because at B the cache moment area is a small enough for us to apply the unit hydrograph theory. So, this d r h can be obtained knowing the intensity of the storm in this area B and then total runoff can be computed at B similarly we can apply the unit hydrograph theory for the cache moment area at point C and obtain total runoff at C which of course, will be quite different from that at point B because the cache moment area may be larger it may have different slope different properties. So, this is runoff at now the problem is that there is a time lag. So, the time at which runoff occurs at point B and the time at which runoff occurs at point C these time origins are not the same for these two graphs. Therefore, we have to do what is known as routing in this case we will have to route the flow through the channel therefore, it is called channel routing in some cases when we route the flow through a reservoir we call it reservoir or a storage routing. For example, in the Nash model we have done some kind of storage or reservoir routing here let us look at what we do once the flow reaches the point B. So, we know what is the inflow at point B the equation which we use now is same as what we have derived earlier that net inflow should be equal to the rate of change of storage. So, d s by d t equal to i minus. So, I will draw only the stream here now point B point C at B we know the inflow because whatever the unit hydrograph theory gives us we add the base flow and get the total runoff and that runoff at point B becomes the inflow in the stream at point B. So, if we look at the stream there is water level in the stream and there is some inflow if we consider a section like this there will be some outflow and the continuity equation tells us that d s by d t would be equal to i minus. So, when we are routing the flow through the channel what we what we need is the outflow for any given inflow. So, basically given an inflow hydrograph like this at point B what would be the outflow hydrograph at point C which we would call which may be something like this this is what we need to find out how will the outflow at point C look because of this outflow at B moving through the channel and reaching point C because when the flow is routed through the channel it will not remain in shape if the inflow hydrograph is like this the outflow hydrograph may look like this i. So, here we are indicating over for outflow i for inflow inflow will have some peak and the outflow will generally have a smaller peak o p. If you look at this curve the area before these two curves intersect represents that outflow is less than inflow and therefore, water is going into storage and what is happening is that this water level is rising and therefore, the storage is increasing. So, water is going into the storage up to this point and beyond this point water is being released from storage. So, we call this as the storage or accumulation and this as release. So, inflow more than outflow means water is going into storage outflow more than inflow means water is coming from the storage or being release from the storage and the aim is of all these routing flood routing or storage routing is to obtain this curve for o given the curve for i and for that we need some kind of relation between s and i and o. For example, in Nash model we have assumed a linear relationship between s and o, but in most cases s will depend on both i and o and therefore, typically we write s as a function of both i and o and there are different methods which we will discuss later on for routing a flood through a reservoir or a channel. Right now we will not go further into this discussion, but this s has to be described as a function of i and o. For example, there is a method known as Muskingum method where s is written as k into some factor x into i plus 1 minus x into o. So, if we assume some kind of relationship like this because the storage will be a function of i the amount of flow coming in will decide what is the water level at this point. Similarly, the amount of flow going out will decide what is the water level at this point. So, the storage typically will be a function of both the inflow and the outflow. The factor x can be taken as 0.5 that means we are giving equal weight to inflow and outflow. If we take the factor x as 0 then we get a linear reservoir as in Nash model which tells s equal to k into outflow. This is called the Muskingum equation and when we route the flow through the channel what we do is we take small time steps let us say delta t. We assume some inflow to be average during the time delta t and then we write delta s over delta t average of i 1 plus i 2 minus average of o 1 and o 2. So, in the finite difference form we write an equation like this. This s delta s will be nothing but s 2 minus s 1 and these s 2 and s 1 again will be correlated with i 1 i 2 and o 1 o 2. Finally, in this equation the only unknown will be o 2 because the conditions at the beginning of a time step are known. So, o 1 i 1 will be known condition at the end of the time step the input would be known because the inflow hydrograph is known to us. So, i would be known at both i 1 and i 2. So, o 2 will be the only unknown which we can solve from this equation. Now, let us come to the third limitation of the unit hydrograph method which we say is lack of data. What we mean by lack of data is there may be a stream or a catchment where there is no gaging station. So, typically we choose this point A as a point where there is some gaging station. So, we can measure the flow at that point and correlate it with whatever rainfall is occurring in the catchment area. We use that information to derive the U H at point A, but if there is no gaging station at point A then we do not know what is the runoff. Even if we have a gaging station at point A it is possible that the required data of given intensity constant duration of storms may not be available to us. So, in that case we use what is known as a synthetic unit hydrograph and the idea of synthetic unit hydrograph is that the shape of the hydrograph will depend on some basin properties. So, if this is the E R H which has unit volume 1 centimeter depth and this is unit hydrograph then this unit hydrograph has some properties or some parameters which will depend on the basin properties and can be correlated with the basin properties. So, one of the important basin property for example, is the length. This will depend on the size of the basin. So, larger the length larger will be the size of the basin. Another important property which represents the shape of the basin is called L C which is the length to the centroid where this point is the centroid of the catchment area and we take a point which is just opposite the centroid on the stream and find out L C along the stream up to the centroid point. This L C will characterize the shape of the catchment area. For example, if you have a catchment area like this its centroid will be closer to the stream and therefore, L C will be smaller. But if you have a catchment area like this the centroid will be further from the point A and therefore, L C will be larger. So, size and shape of the basin can be characterized by these two lengths length L and L C and what is generally done is the lag between the peak of the unit hydrograph and the centroid of the rainfall is typically taken as a function of the basin size and shape and there are lot of empirical relationships based on observed values of these lags and the corresponding length and L C. So, the time to peak the peak discharge are all these factors dependent on the basin characteristics and based on available data lot of these synthetic hydrographs have been developed which help us in plotting this curve by defining the peak time to peak sometimes the width of hydrograph at different locations and so on and we would look at all these synthetic hydrographs in detail in the next lecture. So, in today's lecture we looked at the limitations of the unit hydrograph theory mostly we looked at the three limitations which are lack of data which can be taken care of by synthetic unit hydrographs synthetic unit hydrographs. The area being large for which unit hydrograph cannot be applied that we can take care of by routing the flow through the channel and then if the rainfall is quite non uniform we can take care of that by applying the I U H or instantaneous unit hydrograph theory. So, in the next lecture we would go into details of the synthetic unit hydrograph what are various basin properties which affected how to obtain the peak flow and time to peak we would also look at some methods of routing the flow to obtain the discharge at the outlet.