 We have looked at some methods of finding out the flood for a given catchment. Some of these methods are rational method in which we use a runoff coefficient and some intensity of rainfall and use the discharge as C into I into A where I is the intensity A is the area and C is the runoff coefficient. The other methods which can be used to find out the flood due to a particular storm or the design value are the unit hydrograph method which we have already seen in details how to find out the hydrograph for any given rain if we know the unit hydrograph for a particular duration. The other methods which are normally used are empirical methods which are based on a certain area for which we develop those equations and therefore, they will be applicable only to those areas or similar areas and then we can do flood frequency analysis if we have a gaged catchment and we measure the annual flood let us say for a period of 30 years 40 years based on that we can extrapolate and estimate what will be the flood value for let us say 100 year period or 200 year period. So, we will look at some examples of these methods we will start with the rational method which expresses the flood as q equal to C runoff coefficient I is an intensity of rainfall and we will see how to determine this I and A is the catchment area. Let us say that we have in the catchment which is given the area of catchment is 25 kilometer square. So, the catchment which is given has 25 kilometer square area the other parameters which are given for the catchment are the length and the length to centroid. So, the length along the water course L and then centroid of the area may be somewhere here the point directly opposite that on the water course we can write as L c. So, length of the catchment is given as 11 kilometers and the length to centroid from the outlet point let us call this point A is 7 kilometers. Slope of the catchment is 0.006 generally it is taken as the difference between the points for this away from the outlet point and the outlet point divide by the length L. So, this is also given to us 0.006. Now, we can find out the intensity of rain if we know what duration of rain fall to use and for that purpose we know that as the duration increases the intensity decreases. So, we should take the minimum possible duration and we have also seen that the duration should not be less than time of concentration because then the entire area will not be contributing to run off for some time period. Therefore, that intensity should be taken for a duration corresponding to the time of concentration and we should find out the time of concentration for the given catchment. The time of concentration can be obtained based on empirical equations. For example, as equation which is similar to Snyder's equation in the Snyder's synthetic hydrograph we have the time to peak is equal to some constant C t into L and L c which are the length of the catchment and the length of centroid. This factor was not there originally in the Snyder's equation, but we use this square root of S term. This is the square root of the slope of the catchment. This is L into L c and then this is the power n. Again, the C t and n value they differ from catchment to catchment, but for this catchment let us say that we are given these values as 0.5 and 0.27. So, using a relationship similar to Snyder we can estimate time of concentration by assuming it to be equal to the lag and for small basins this assumption may be valid and if we use this equation L and L c of course, in this equation are in kilometers slope is dimension this and the value which we get will be in hours. So, we get a time of concentration of 3.2 hours which means that a rain drop falling here or here which is the farthest really from a travel point of view will take about 3 hours to reach the outlet A. Now, this is one equation which we can use to find out time of concentration. There is another commonly used equation known as the Kerr-Pitch equation which says that T c in minutes will be equal to 0.01947 into this L in meters and this is the slope which is dimensionless. So, the Kerr-Pitch equation says that T c in minutes will be 0.01947 times the length of catchment in meters to the power 0.77 divide by the slope to the power 0.385. If we put the values of L and S we get a value which is 181 minutes. So, if we look at these two in this case they are not very different. So, to be on the conservative side we can take a T c which is smaller than this and therefore, we will use a duration of 3 hours. That means for rainfall duration of 3 hours we have to find out the intensity and as we have seen earlier there are equations which relate the intensity with the duration and the return period. So, one of the equations which we had looked at was of this form i equal to intensity equal to k which is a factor depending on the catchment T is the return period in years x and n are exponents T is the duration in hours and a is another constant which depends on the catchment. So, assuming these values for k x a and n we can find out for a return period of 25 years what will be the intensity of a 3 hour rainfall. So, we put the value of k as 100 millimeter per hour the exponent x as 0.2 and n as 0.9 and then a as 0.5 hours using d of 3 hours we get a intensity of 62 millimeters per hour. So, in the relationship q equal to c i a i is obtained based on the time of concentration and the catchment properties a is given as 25 kilometer square. Now, we have to find out runoff coefficient we have already seen that runoff coefficient depends on the type of area. In this case let us say that this 25 kilometer square area has 3 different types of areas or land uses. So, here we have listed the values of residential area is 6 kilometer square agricultural can be taken as 17 and there is 2 kilometer square of paved area. The runoff coefficient c for these areas is given in tabular form in various books. So, we can take the representative values of c as 0.5 for residential 0.25 for agricultural agricultural it depends on the soil also in this case we may assume it to be a little sandy and therefore, runoff coefficient c will be a little smaller. So, 0.25 is and then paved we can use a value of 0.9 the runoff coefficient is generally very high for paved areas closer to 1. So, we will take a value of 0.9. Now, we have to find a mean value of c for the entire catchment. So, we can give a weight which corresponds to the area which is occupied by the corresponding land use and therefore, we can write a weighted mean of c as 0.5 for residential area 6 area occupied by residential 0.25 again c for agricultural agricultural area paved c and paved area divided by total area of the catchment which is 25 kilometer square. So, we get a weighted value of c as 0.36 and now we can obtain the value of q 0.36 is c 62 is in millimeter per hour. So, we will convert it into meters by dividing by 1000 and per second by dividing by 3600. 25 is the area in kilometer square. So, we multiplied by a million to get in meter square and therefore, the resulting value will come out to be in meter cube per second and it turns out to be 156 meter cube per second. So, over this catchment of area 25 kilometer square if a rainfall occurs of intensity 62 millimeters per hour for 3 hours then the maximum discharge which can be expected will be 156 meter cube per second using the rational formula. There are empirical equations also which can be used to find out the flood discharge. We would look at some of them. For example, for the same catchment area of 25 kilometer square we have seen these equations which are given for a particular area. For example, Dickens equation is northern India, hilly regions or central India. The value of c is different for different areas and we will use let us say c of 15 for this case which typically is the higher limit for hilly area in northern India or in central India lower limit. So, let us use 15 area in kilometer square to the power 0.75 gives us a value of 168 meter cube per second. Similarly, there are other equations like Rive's equation, Anglia's equation which are derived for Tamil Nadu or Maharashtra particular area they are derived for they should be used only for that area or similar areas, but if we use them using some coefficients we can get an estimate of the flood discharge. In this case 87 English gives us 521 which is very high and then in the US the Fuller's equation is commonly used which accounts for the return period also. So, these other equations they do not have any return period term included there, but Q here for Fuller's equation accounts for the return period and using a c of 1.9 which is on the higher side we get a value of 52 meter cube per second and this is a 24 hour flood. So, using empirical equations we can obtain a 24 hour flood or flood for a any given duration, but typically these are giving us daily flow daily floods not the 3 hour flood which we obtain from the rational method. So, we would then look at the frequency analysis in which suppose we have a gaged catchment and we have the data of the flow available for let us say daily flow data we have. So, again suppose we look at the same some catchment area A and suppose we have this gaging station here. So, Q is known for measured now from this measurement we can get for each year what is the maximum flood and this table shows you the year versus maximum flood and using this data we can do a frequency analysis. Note that these 2 values are same. So, when 2 values are same when we rank them we will have to be careful. So, what we do first in the flood frequency analysis is arranged data in decreasing order. So, this data which corresponds to 25 year period from 1976 to 2000 can be arranged in decreasing order and given a rank. So, in this table we are showing the rank from 1 to 25 Q is arranged in the decreasing order and here you will notice these 2 values 47 34 are the same. So, the first value is not assigned any return period. This return period in this case n is 25 because we have 25 years of data. So, the return period would be n plus 1 divided by m or in this case 26 divided by m. So, for the first it will be 26 divided by 1 26 divided by 2 and so on. So, we can compute the return period for each flow and since this return period represents the flow being equal or exceeded when the 2 values are same we assign it to the lower value in this case or the higher rank value 13 and get return period of 2 for this case. So, for this case the value which we will actually obtain would be 26 by 12, but we will not write it here because these 2 are the same and they both will have a return period of 2 years. So, this table completes up to 25 this is the minimum annual maximum flood which has been observed and this is the maximum you can see it goes from about 3500 to 7000 mean of these flows is 4889 and the standard deviation we will be using these 2 in the probability distributions later on. So, we can compute the mean of the queues and the standard deviation which is so mean let us write as mu of q and standard deviation sigma of q mean of course, is sum of all sigma q divided by n and this is based on square root of q minus mu q square divided by n minus 1. So, that formula gives us the values of mean and the standard deviation. Now, there are various frequency distributions which have been used for annual maximum flood we will discuss 2 of them one is the Gumbel's method and other is the log Pearson type 3 distribution. So, we start with the Gumbel's method in which the reduced variable y is dependent on the actual variable x or in this case q as 2825 x minus mu x divided by sigma x plus 0.577. So, a reduced variable has been defined which has written period of t. So, t again in this case is the written period. So, for any particular q we can obtain the written period which is the same as we have shown here this q 6978 has written period of 26 and for this written period we get the reduced variate using this equation. And therefore, this reduced variate versus q plot can give us a method of extrapolating the value of q for a higher written period. So, in this case we have plotted here the reduced variable versus q notice that this q is in 1000 meter q per second. And so, for example, for 6978 the reduced variate comes out to be 3.24 which corresponds to written period of 26 years. So, that 6978 point is shown here with the written reduced variate of 3.24. Similarly, if we look at reduced variate of 1 q is 5212. So, that corresponds to this point. So, all the points can be plotted and in this case if the data follows the numbers distribution they come on the straight line very nice straight line which can be extended to extrapolate the value for any written period. For example, if we want 100 year written period for example, we want 100 year plot. Then we can find out the reduced variable corresponding to that t using this equation which comes out to be 4.6. And then using this data at 4.6 we would estimate what is the value of the 100 year plot. So, 4.6 will be somewhere here and then corresponding to this we can estimate the value and obtain it graphically from this comes out to be about close to 800 meter cube per second. So, we can say that the 100 year flood is roughly 8000 meter cube per second. Most of the times the data is not infinite because gumbo's distribution the values which have been used there they were derived for an infinite set of data. But if we have finite set of data then we need to make some corrections in the method. So, if we have a finite data set in this case since we had only 25 year of data we must modify the equations. So, this is the equation which we use saying that this factor k which depends on the reduced variable y mean and standard deviation this mu x we have already obtained from the data. But now this k instead of having a value y n as S n which were earlier given as 0.577 and 1.28 instead of these values now they will depend on how many years of data we have. So, from the table which is again given in various books we obtain a value of y n for 25 year data as 0.5309. If we have a large number of data set then 0.577 is the value. So, there is some change here similarly S n for infinite data set 1.28 and for 25 year required 1.09. So, this gives us a value of k let us say we want 100 year flood reduce variable we have already seen will be equal to 4.6. So, 4.6 minus 0.5309 divided by 1.0915 will give us a value of k equal to 3.728. And then using this equation where mu is known sigma n minus 1 is known k is also obtained from here we can get the 100 year flood as 8126 meter cube per second which is very close to what we had obtained graphically at about 8000 meter cube per second. So, for finite data set we need to correct these tables will be available in various references or books and we need to correct it and then obtain the value of the flood. So, in this case we obtain 8126 earlier we had obtained 8000 meter cube per second they are very close to each other. Now in Gumbel's method since it is a linear relationship if we know the flood for any two return periods we can extrapolate for any other return period. So, we will take an example in which from two values we can extrapolate to any other value. So, the example which we will take is let us say that the given data is for 100 year flood and 150 year flood we know what is the flood discharge and using that we find the we have to find the 500 year flood. The same thing can be done using the graphical technique also in which we would extend this line and find out for 500 what will be the reduced variable y using this equation and then corresponding to that y we can find out what is the flood, but using computations also we can estimate that. For example, let us say that q 100 is given as 8000 meter cube per second and q 150 is given as 8400 meter cube per second. So, on an average 8000 meter cube per second of maximum flood annual flood can be expected once in 100 years 8400 meter cube per second expected once in 150 years and what we want is what will be the flood which can be expected once every 500 years. So, for t equal to 100 we have already seen y is 4.6 for t equal to 150 we can obtain y as 5.0 and for t equal to 500 y is 6.2. So, using the 6.2 we could have extrapolated this value and obtained say 6.2 will be somewhere here. So, we can see that it will be close to about 9500 or so, but using the calculations also we can look at this how to compute this. The equation which we already have seen is that q for any return period t will depend on the mean of the discharges. So, mu x which represents the mean of the discharges and sigma which has standard deviation these two are constants they are dependent on the data set available they do not change with the return period k changes with the return period and as we have seen already the equation for k is y t minus y n divide by s n. So, y n s n sigma n minus 1 and mu x these values they do not change with t the only thing which changes with t return period is the value y t. So, we can write this as some constant a plus some other constant b into y t because everything else is constant except y t and it is a linear function of y t. So, if we write q t equal to a plus b y t we have really two unknowns a and b if the values of q t and y t are known for two different values then we can estimate these a and b values in this case q 100 is given and y 100 is given similarly, q 150 is given and y 150 is given. So, using these two we can obtain b as delta q divide by delta y which comes out to be 1000 meter cube per second. So, 8400 minus 8000 divide by 5 minus 4.6 will come out to be 1000 meter cube per second and similarly a can be obtained from any of these equations as 8000 minus 4.6 into 1000 and it turns out to be 3400 meter cube per second and then using the value of y for 500 years we can obtain q 500 as 3400 plus 6200 which is 9600. So, it comes very close to the value which we obtain from the graphical method, but this can be obtained directly without the need of plotting the graphs. So, a 500 year flood for the given data can be taken as 9600 meter cube per second. So, using the frequency analysis we can obtain estimate what is return period what is the flood for a particular return period, but since these are all uncertain values we should have some idea about not only our predicted value of flood for a return period, but what is the confidence level. So, typically we will analyze let us say 95 percent confidence interval. So, we will predict a range of discharge values and say that there is 95 percent chance that the actual 100 year flood or 500 year flood would be within those values. So, we will look at the confidence interval in this case the confidence interval is given as what is the value we predict for T year return period and then there is a plus minus. So, there is a range about the mean or the predicted value within which we have a 95 percent or 80 percent confidence interval f alpha is a function of the confidence interval. So, for example, for alpha equal to 0.95 we get f of 1.96. So, there is a table of values alpha versus f alpha which is again given in various references and from that we can pick up the value of f for any given alpha S e is a probable error which is given as some constant and that constant is related with k and this square root sign is over this whole term. So, this term is a function of k then we have this standard deviation and square root of n, n is the number of data in this case we have 25 k can be obtained using the same equation for T equal to 100 years it can be obtained as 3.728 with given values of n and sigma. So, using this data we can obtain the probable error using k we can obtain this constant this is known this is known. So, we get the value of S e as almost 800 meter cube per second x t which we have estimated 100 year flood earlier was 8126 meter cube per second. So, if you want to find out the 95 percent confidence interval we will use this equation with x t of 8126 f alpha of 1.96 S e of 798 and we get 8126 plus minus 1564 meter cube per second. So, this gives us a range for 100 year flood. So, what it tells is that 100 year flood will be between 65, 62 meter cube per second and 96, 90 meter cube per second if we make this a statement we are 95 percent confident that the data actual flood will lie within this range. So, instead of predicting a single value which earlier we had taken as 8126 we have now predicted a range. So, this means that the actual flood will be between 65, 62 to 96, 90 there is a 95 percent probability that the actual flood which will occur once every 100 years will be within this range. Now, if we just want to have a different probability or confidence interval for example, if we take alpha to be 0.9 then the only thing we need to change is f alpha which comes out to be 1.645. So, when we put that we will get a range here which will be smaller than this range because f alpha has become smaller. So, a 90 percent confidence interval would be smaller about again it will center on the mean 8126, but the band will be smaller. So, as we reduce the confidence interval the band will get narrower and narrower, but 95 percent confidence interval is typically used and therefore, in this case we can say that the design value of flood should be between these values. So, we can estimate the flood and we can also estimate the confidence interval that within that interval the flood is likely to be 95 percent certain with 95 percent certainty. The other distribution which we use is known as the log Pearson type 3 distribution this is commonly used in the US, where Pearson type 3 is a distribution which is assumed and log means that instead of the variable following the Pearson type 3 distribution log of the variable follows the Pearson type 3 distribution. So, we use a variable z which is defined as log of q and then we say that z will follow a Pearson type 3 distribution and for any return period t z value will be given by mean again a constant k and the standard deviation sigma z. Now, in this case the distribution is supposed to be skewed. So, it is not a symmetric distribution like this, but it can be skewed like this or it can be skewed like this. There may be a long tail in the negative side or there may be a long tail on the positive side. So, that will give us some skewness and the function k z which we have here k z is a function of the return period t and the skewness of the distribution which we call c s. The c s value is computed by using this equation where z minus mu z cube will be 0 if the distribution is symmetric because if it is symmetric then positive and negative values of the cube this is cube positive and negative values of the cubes will cancel each other and c s will come out to be 0, but typically it will not be. So, for example, in this case we can compute c s from the given data. There are tables given for c s. So, for example, for a 100 year flood if c s is 0.2 k z is 2.472. If c s is 0.3 k is 2.544. We have taken these two values of c s because when we compute c s for this data we find it to be between these two. So, that is why from the table these two values we can take. The data is shown here the same data 1976 to 87 this figure and then up to 2000 this figure. So, from this data we first transform it this is log of q. So, we transform the q to its log then this is the difference of square so z minus z mu z square and this is cube z minus mu z cube. So, we find out the squares and the cubes for all the data and then we add them up mean of course, this is the mean of the log q not q. So, it comes out to be 3.683 standard deviation from the summation of these squares divide by n minus 1 square root will give us 0.0753 and skewness c s is obtained from the equation shown here where this z minus mu z cube is the sum of the column and sigma z we have already obtained here. So, using this sum and sigma z and mu z we get a value of c s as 0.275 corresponding to this c s as we have seen here in this table k is given for 0.2 and 0.3. So, we can use the linear interpolation between these two values and obtain the value of c s k value for the given c s as 2.526. So, using this value of k we can estimate z for 100 year written period as the mean k and sigma mean and sigma we have already seen from the table mean standard deviation k we have obtained here. So, the c 100 comes out to be 3.873 and q 100 is nothing, but 10 to power z. So, in this case it will come out to be about 7500 meter cube per second. So, using the Gumbel's distribution we had obtained a value close to 8000 here we are getting almost 7500 meter cube per second. So, both of them are quite close. Now, sometimes since it is a finite data set the c s value is adjusted has not has proposed this modification in which the c s is modified by this factor of 1 plus 8.5 over n, n is the number of years of record in this case number of data set is 25. So, 1 plus 8.5 divided by 25 into c s will give us value of 0.37 for this value again from the tables we can obtain the value of k z 100 and q 100. So, in this case it becomes little more than this value, but still close to 7500. So, using either the Gumbel's method or the log Pearson probability distribution we can obtain the value of the flood, we can obtain an estimate of the confidence interval also. The next thing which we can do is we can find out what is the risk or reliability of a structure. So, for example, if we take suppose we design a structure which is of a useful life of 50 years and we design it for 100 year flood, what will be its reliability or what is the chance that it will not fail in the next 50 years. So, that is the what we call the reliability probability of the event not occurring in n years where n is the design period. So, what we say is we have designed a structure for let us say 100 year flood and the useful life or the design life of the structure is 50 years, what will be its reliability. So, in this case we can find out 1 by t is the probability of the event occurring. So, 1 minus 1 by t will be the probability of event not occurring in 1 year and to the power n will give us the probability of the event not occurring in n years n consecutive years. So, in this case using n equal to 50 and t of 100 years we get a reliability of 0.61 which means the risk which is 1 minus reliability that is the probability of the event occurring at least once in n years or this is the probability of failure of the structure will be 39 percent. Now, 39 percent risk may be too high in some cases. So, if we want to say that we want to keep a risk of let us say 10 percent or 5 percent then we can find out the corresponding return period. For example, if we say that the risk is to be reduced to 20 percent what should be the design discharge for what return period. We can use the same equation 1 minus 1 by t to the power 50 equal to 1 minus risk equal to 0.8 which gives us a t of 225 years meaning that if we want to have a 20 percent risk only then we should take the 225 years flood or 225 years event to design that 50 year structure. So, we have seen the risk reliability there is another term which is also commonly used to define the safety of a structure safety factor and safety margin. So, these are quite a straight forward that suppose we have 100 year flood of 8126 meter cube per second and the design value which is used is 9000 then the safety factor will be simply the ratio of the actual value to the estimated value. So, the used value is 9000 estimated value of the variable is 8126 the we have a safety factor of 1.11 and sometimes we use safety margin which is the difference of these two. So, we have a safety margin of 874 meter cube per second. So, we have looked at how to estimate the design flood how to obtain the reliability or the risk. So, in all these cases we have assumed that there is some method of estimating the design flood at that location where we want the structure to be built it may be a dam or a bridge. In some cases the flood may not be available at a particular location or in some cases we may have to design a structure by passing that flood through that structure. For example, in case of a spillway we have a flood coming in there will be some storage and then there will be some outflow. So, we need to know what will be the rise in the water level when the flood comes because the inflow and the outflow will not be same due to this storage effect. In some other cases suppose there is a river at one point we may be knowing the flood hydrograph and then we want to estimate what will be the flood hydrograph at another location downstream of that point. So, this is known as the routing or the flood routing which typically we will discuss in terms of storage routing where the flood is being routed through a reservoir. So, storage routing or reservoir routing or level pool routing because we assume that the water level in the reservoir remains horizontal and sometimes we would also look at channel routing in which there is a flow coming in a river or a channel and then how it will move downstream. So, the routing the basic aim is to find out how the inflow hydrograph gets changed as the outflow hydrograph because outflow hydrograph will be different from the inflow hydrograph and we need to find out how it will be different. So, for example, if we have let us say a dam here and there is some inflow coming in this water level here the discharge over this will be q will depend on what is the water level here or what is the height h. The storage within the reservoir will also depend on the height h and i as a function of time which is the inflow hydrograph will be given to us. So, given this inflow hydrograph we want to estimate how this water level will change in such a way that is storage and outflow they account for whatever inflow is coming. The equation which is used for this case is the continuity equation which says that in any time period the net inflow should be equal to the change in storage. So, if we have some time t equal to t 1 and t equal to t 2 then delta t during that time there is mean inflow and mean outflow because inflow is changing with time. So, we can use some mean value between t 1 and t 2 typically it is done as i 1 plus i 2 by 2 where i 1 and i 2 are the inflows at the beginning and at the end of the time period. And these will be known because inflow hydrograph is known to us q similarly q bar is taken as q 1 plus q 2 by 2 q 1 is known to us at the beginning of the time step, but q 2 is not known to us similarly delta s is s 2 minus s 1 s 2 and q 2 these are the two unknowns. They are both functions of q functions of h and our aim of the routing is to find out how q and s and h they change with time. The method which we use in this case will be the modified Holtz method uses the same form i minus q mean value delta t equal to delta s which is converted in terms of i 1 plus i 2 by 2 delta t known quantity plus another known quantity s 1 minus q 1 delta t by 2 this is equated with the unknown which is s 2 plus q 2. So, what we do is this q contains q 1 and q 2 q 2 is not known. So, we transform the equation in such a way that the known quantities come on one side and the unknown on the other side. Then suppose in this case we assume that the storage above the reservoir level at any height h the storage above the reservoir level can be given by h plus h square million cubic meters just for our computational sake we will assume this equation, but typically it will be given in the form of a graph as h over s and generally the form of the equation will depend on the topography of the area and the nature of this variation may be linear may be quadratic. So, in this case we have assumed a quadratic distribution s as h plus h square similarly, the outflow q will also depend on h and we have assumed generally it varies as h to the power 3 by 2. So, we have assumed that q is equal to 100 h to the power 1.5 meter cube per second. So, suppose this data is given to us that s and q vary with h as these given functions the inflow hydrograph is given at 4 hour intervals and now we want to route the flood using a delta t of 4 hours. So, if we look at this table we have prepared a curve or we have obtained data which relate s minus q delta t by 2 and s plus q delta t by 2 with h. If we look at the pulse modified pulse method equation this is s minus q delta t by 2 s plus q delta t by 2. So, both these quantities can be related with h because s and q both are functions of h for any given delta t which we have assumed here as 4 hours. We can prepare a curve between h and these 2 quantities and that is what we have done here we have computed s minus q d t by 2 and s plus q d t by 2 versus h. So, for any given h s is obtained as h plus h square q is obtained as 100 h to the power 1.5. We have taken a range up to 0.7 we could go higher also, but in this case we have stopped at this because the outflow does not go above this value as we will see later. And obtaining these 2 columns we can plot it and show the variation of h versus s plus minus q delta t by 2. So, s minus delta t by q delta t by 2 and s plus these 2 curves can be plotted and pulse method consists of graphically doing the computations. So, when we start i 1 plus i 2 by 2 delta t we can compute s minus q delta t by 2 we can obtain from the graph for the given h. In this case we will assume that when the flood starts the inflow hydrograph starts at t equal to 0 and at that time the water level in the reservoir is at the top of the spillway. So, this is at time t equal to 0 we have this is the pool level and therefore, q is 0 initially let us assume that q is 0 we could assume some other value also, but let us just assume that since the water level is here there is no outflow we could have assume the different water level also. Now s minus q delta t by 2 will be known for the starting value of h i bar delta t we can obtain from the inflow hydrograph. And therefore, we can get this value of s plus q delta t by 2 and once we know this value corresponding value of h can be obtained from this figure. So, the calculations proceed like this t in hours and inflow in meter cube per second is given to us s minus q delta t by 2 0 because our initial h is assumed as 0 h 0 is assumed as 0 therefore, both s and q are 0 this term comes from s minus q delta t by 2 plus the mean inflow into delta t. So, 28 plus 10 divided by 2 which will be 19 meter cube per second into 4 hours that we will convert into million meter cube and get a value of 0.274 corresponding to this 0.274 we would look up from the graph and see the value. So, 0.274 would be somewhere here and from the graph we can look up the value of h it turns out to be around 0.183. So, we start with 0.274 get 0.183 now for this 0.183 we can obtain the value of s minus q delta t by 2 again from this figure. So, s minus q delta t by 2 for the next step will turn out to be about 0.16. So, that is what we have done here in this value we will add again the mean inflow during that time period. So, add this and we will get a new value of s plus q delta t by 2 as 0.852 and for 0.852 again we can get the corresponding value of h as 0.443 from this figure then for 0.443 we can get the value of s minus q delta t by 2. So, for 0.443 we get 0.427 as s minus q delta t by 2. So, basically we just go from one curve to the other. So, we come to this point get to this curve then similarly once we obtain the value here we come to this point and get the value for s minus q delta t by 2 then again add the mean inflow to get s plus q delta t by 2. So, in this way by doing the computations we can obtain at the end of each time interval what will be the value of h and therefore, q because q is a function of h we have already seen and we can plot this time versus q as the outflow hydrograph which is shown in this figure this is in inflow i and outflow. So, a few things can be noted from this figure one is that at this point where the inflow and the outflow cross that means inflow is equal to outflow at this point the outflow is also maximum. And as we have discussed earlier that it will be maximum because at this point the storage is at the maximum before this water is going into a storage after this it is being released. So, this much water goes into the storage at the this point water level reaches a maximum because q is also maximum storage also maximum and then beyond this water will be released from a storage and therefore, outflow will be more than the inflow. The peak of the outflow as we can see from this figure this table is about 55 meter cube per second the peak of the inflow is around 68. So, there is an attenuation or a decrease in peak of about 13 meter cube per second and it has also shifted. So, the peak shifts or there is a lag and there is an attenuation in the outflow hydrograph. So, in this way we can perform the storage routing and obtain the outflow curve and also how the water level in the reservoir is changing with time that also we can obtain. So, we can see that initially it starts with 0 then it increases up to about 0.668 and then it starts again decreasing and in this case it has gone down to 0.235 it will go down further if we continue the computations. In the storage routing we have the storage dependent on head in channel routing which we will now see channel routing means if there is a flood coming in a channel there is some depth of flow. There is really not a very defined storage in this case, but there is a change in storage and we have already seen that we can write this storage as wedge plus prism. So, if we have this change in water level like this then we have a prism storage and a wedge storage and we will be discussing the Muskingham method of routing. In the Muskingham method we assume a linear reservoir or linear storage as s. So, the storage in the channel the wedge storage is proportional to the inflow the prism storage is proportional to the outflow and the storage channel storage total storage is given by sub factor k which is the time constant some weightage x to i and 1 minus x to q. In this case let us assume that the values which are given are 8 hours for k 0.2 for x and let us use same inflow hydrograph with the d t of 4 hours. There is computationally there is a requirement that d t should be between 2 k x and k. In this case k is 8 and 2 k x is 3.2. So, we use 4 hour which is good for this case inflow hydrograph as we have seen is given at 4 hour interval. Now, in the Muskingham method the equation which is used basically we use the same equation, but the change in the storage is written as i 2 minus i 1 plus 1 minus x q 2 minus q 1 and using this storage we can write an expression relating the q at the end of a time period with inflow i 2 at the end of the time period i 1 at the beginning and q 1 at the beginning. So, these q 1 i 1 and i 2 are known c 0 c 1 c 2 are constants which are obtained by these relationships and the numerical values of these constants is obtained as 0.0476, 0.429, 0.524 and therefore the computations are rather straight forward in this case for the time given inflow is given and q 2 is obtained from this equation where c 0 c 1 c 2 these are all known i 2 i 1 and q 1 these are obtained from this. For example, in this case i 1 is 10 i 2 is 28 and q 1 is 10 these 3 values will give us 10.86. When we go to the next step then we use i 1 i 2 q 1. So, these 3 values will give us in this way we can proceed our calculations like this and get the outflow hydrograph which is plotted here. In this case we have assumed that initial outflow is 10 meter cube per second because this is channel routing. So, we assume that initially inflow and outflow are same and therefore again you can see that there is an attenuation here and there is a lag here, but in this case the peak is roughly 54.11 while in the other case we had seen 54.62. So, they are not very different in this case. So, we have seen how to move the flood or how to route the flood from one point to the other if the inflow hydrograph is given at one point and if the storage characteristics of the channel or the reservoir are known then we can obtain the flood hydrograph at any point downstream.