 In the previous few lectures, we have seen about the floods, how to estimate the magnitude and how to obtain the value of discharge at different times at a particular location in let us say catchment area. Now, once the flood is obtained at the outlet from a catchment area and suppose we want to predict the flood at a downstream location, we will have to find out how this input gets translated into the output at the downstream point. And we use for this purpose what is known as routing. The flood routing is an important subject because it will translate any input at one point to the flood output at a downstream point by taking care of the storage resistance and other factors. So, today we will discuss some aspects of flood routing. Flood routing simply means that we route the flood from one point to some point downstream. So, we translate the flood and obtain temporal distribution, how the discharge is changing with time of let us say discharge flow generally meter cube per second with at downstream points. So, for example, if we have a channel and we know the temporal distribution of discharge, let us call it the inflow and at this point we want to obtain temporal distribution of the outflow which we can call q, how will it will it be like this or will it be like this. So, the purpose of flood routing is to obtain or estimate this distribution of q. Hydrograph at one or may be more points there may be a tributary coming in here and the hydrograph in this tributary may also be known. So, at a few points upstream of the point at which we want to find out the hydrograph using these points we can estimate the outflow at this point using this techniques of flood routing. The basic techniques used in flood routing can be called either a lumped routing or distributed. In the lumped routing we obtain q as a function of time at a given x. So, for example, again if we take some catchment area there may be tributaries entering this channel. So, for the tributaries suppose we know the hydrographs and all these hydrographs are routed to the outflow point. So, at this point if we find out outflow versus time then we are considering only one point at particular x. So, given x would be this point and what we want is q versus time. This would be a lumped routing in distributed we find out q as a function of x and t. So, at different times in a channel at different x values what will be q. So, for this entire channel we can say that at any time what will be the q at different locations. So, lumped and distributed are kind of have the same purpose of routing the flood, but they use different philosophies in lumped routing which is also called hydrologic routing. We use a continuity equation which is generally of the form inflow minus outflow equal to rate of change of storage. So, the lumped routing or the hydrologic routing uses the simple mass balance equation of inflow outflow and the rating them with the change in storage. While in the distributed model which is also known as hydraulic routing we use a continuity equation and a momentum equation. So, for example, if you are talking about a channel and the depth of flow is y at different times we can write a continuity equation which would be of the form and q is the discharge q will be a function of x and t. The continuity equation can be written as this is the change of discharge with distance this denotes the net inflow or outflow into a particular control volume of very small with delta x and then in the limit that delta x tends to 0 we would get the differential equation. T is the top width. So, if the channel cross section is like this and water is flowing at this level then t would be the width at the top of this section y is the depth of flow and q is the discharge. So, this gives the continuity equation relating the net inflow with the change in control volume within this control volume. Momentum equation is obtained and written as del y by del x v by g del v by del x s naught minus s f. S naught is the bed slope of the channel. So, if the channel is here the slope of the bed will be s naught and s f is the friction slope. The friction slope represents the energy loss. So, if the channel bed is here water depth may be like this the energy line may be going like this. So, this is the water level and this represents the total energy or energy line which is obtained by adding the velocity head. So, the slope of this line at any point is generated by s f and typically this value of s f is obtained from using equations like Manning's equation or Chagey's equation for open channel flow. So, in this equation s naught and s f are obtained from bed slope is obtained from the geometry of the channel s f is obtained from Manning's or Chagey's equation and then we can solve these two equations together to obtain the value of discharge at all values of x and at different times. So, we will start with the lumped routing because it is easier. Now, all the routings can be thought of in terms of two different kinds either storage routing or a channel routing. The storage routing is also called reservoir routing and as the name implies typically it denotes how a flood wave gets modified when it passes through some storage structure or reservoir. So, let us look at a case like this where we have some storage reservoir here water is stored up to certain level and there is some flow coming in I t is the input and then the outflow can be written as let us say q t and we want to find out how the outflow is changing with time. A storage in the reservoir denotes let us denote it by s, s will of course depend on the height of flow in the reservoir. Let us call h from the top of the spillway so that when h is 0 there is no outflow, but we can relate this h we can take the h from the bed of the channel also, but for this our purpose we will take h as the variable. So, s will be a function of h as we go higher the storage volume within the reservoir increases. So, s will typically be a function of h where h equal to 0 indicates the storage below this level and then it increases as we increase the h. Similarly, the outflow from the reservoir will also depend on h and generally if we have uncontrolled spillway here then q typically is proportional to the 3 by 2 power of h and we can plot a curve like this showing the variation of q with h. So, the equation which is used for lumped routing i minus o inflow minus outflow equal to d s by d t this s will also be a function of h and o which we have denoted by q here will also be a function of h. So, what we need to do is to solve this equation and see how outflow is changing with time there are many methods of solving this. We start with a method which is known as level pool routing level pool routing because we assume that reservoir level is. So, we will ignore any changes in the water level any profile occurring like this we assume that the water level at all the times in the reservoir remains horizontal and this assumption will be true or approximately true most of the time because the slope of the water surface here will generally be small. So, assuming this the reservoir routing or sewage routing or this level pool routing we write the equation the familiar mass balance equation as i minus o equal to d s by d t and then we consider a small time interval delta t. So, small interval denoted by delta t and let us see what happens during this small interval finite time interval of delta t the inflow which comes in can be written as some average inflow because i changes with time. And therefore, during the finite time interval of delta t we can write some i bar as the average inflow minus similarly, we can write some average outflow and now let us use the symbol q for the outflow. So, some q bar again multiply by delta t will give us the change in storage delta s. So, in this equation i bar represents the average inflow during time period of delta t q bar average outflow during the time period and delta s is the change in storage during that finite time interval of delta t. We can write this i as i 1 plus i 2 by 2 similarly, q 1 plus q 2 by 2 where i 1 and i 2 represents the inflow before that time period of delta t and after the time period of delta t. So, if t versus i is known to us and suppose we have taken this delta t then this will be i 1 and this would be i 2. So, obviously delta t has to be small enough. So, that we can assume that this portion is more or less a straight line and therefore, this assumption of average flow being equal to i 1 plus i 2 by 2 will be valid and delta s will of course, be the change in storage s 2 minus s 1. So, using this equation now we would like to route the flow through the reservoir and see what kind of outflow we get for a given inflow. The information available to us would be the inflow curve. We also should know how the storage changes with h and how the outflow changes with h of the storage and similarly, q as function of. So, if we have a storage reservoir here initially we should know what is the level let us call it h 0 and then the inflow comes and we want to route it through the reservoir the inflow hydrograph is given as this. So, at time t equal to 0 we know what is the inflow i 0. In fact, the inflow hydrograph is known for all times. So, at any time we can find out what is i from this curve. So, if we take this time step we can use different methods to solve this equation. So, let us write this equation again here, but now in terms of unknown quantities on one side and the known quantities on the other side. So, the same mass balance equation is now written in a slightly different form by putting all the unknown quantities s 2 and q 2 both of these are unknowns i 1 s 1 and q 1 they are at the beginning of the time step. So, these are known and as we said i 2 will be known because this entire curve is given to us. So, the inflow hydrograph is completely known therefore, i 2 will also be known to us. The problem now is that this s 2 and q 2 both are functions of h. So, we cannot directly solve for these in a analytical form. So, there are some graphical methods which are available for solving this equation. The commonly used method called the modified pulse method looks at plotting the curve in a little different form. So, that we are plotting s plus q delta t by 2 versus h. So, if you look at this equation the term s plus q delta t by 2 will be a function of h because both s and q depend on h. The plot of s plus q delta t by 2 would look like this on the same graph we can plot q also. So, this will be on a different scale s plus q delta t by 2 of course, delta t we have to decide. For deciding delta t normally it is selected in such a way that the time to peak typically consists of about 5 3 to 5 delta t. So, this time to peak delta t should be roughly about 0.2 to 0.4. So, that is the general criteria for choosing delta t. Once we choose delta t then since we know both s and q as functions of h plotting this curve is straight forward for a given value of delta t. Now, if we look at the equation all these quantities are known at the beginning of the time step. So, with these quantities we can find the sum of s and q delta t by 2. We cannot find individually s 2 and q 2, but the sum is known to us because of this equation. So, from that known value and from this curve we can directly find out h which gives us the head or the water depth at the end of the interval delta t. Now, on the same figure we have this plot of q on this axis. So, this is q and this is s plus on the same graph we have this plot of q and therefore, for this h we will know what is the value of q. So, for the next time step where we need s minus q by q delta t by 2 we can write this in terms of the known variable minus. So, knowing the value of q from here we can find out q delta t also at the end of the time step delta t and since s plus q delta t by 2 is also known at the end from the graph we can subtract it and get the value of this variable at the beginning of the next time step. So, we solve for one step and then for the next step this will become the known value here and it will knowing the inflow we can again get the parameter s plus q delta t by 2 at the end of time step and repeat this process till we route the whole inflow hydrograph through the reservoir. So, this modified pulse method using this graphical procedure gives us the outflow the typical outflow. So, if we have this as the inflow a typical outflow will look like this. So, t this is inflow and outflow where this shows the inflow and this shows thing to notice here is that till the intersection point the inflow is more than outflow and what it means is that the water is going into the storage or the storage is increasing. And then beyond this point the outflow is more than inflow which means that water is being taken out of the storage and the volume of these two should be same because whatever additional water we have deposited into the storage here should come out during this time. And after this the inflow and outflow will be same at the initial level. So, if they reach the initial level again then they will maintain at that initial level the storage will return back to its original position. The two things which should be noted here one is the change in peak discharge this is called attenuation or sometimes peak attenuation means what is the reduction in the peak as the inflow passes through the reservoir. And the other thing which can be seen from here is the lag. So, this is by how much time the peak has been delayed. So, this is important implication in the sense that the flood at the inflow point is occurring at some point in time. But it will have some lag as it passes through the storage reservoir. So, the peak will be reduced and there will be lag also. So, there will be an additional time to peak which downstream locations will get. So, these two are important from in the sense that how much peak is reduced and how much extra time we get will affect our preparation for protecting against flood on the downstream side. So, storage reservoirs thus serve these two important purposes that they reduce the magnitude of the maximum flow or the peak and they also lag the peak discharge by some time. There is another method or sometimes what we do is we prepare another curve s minus q delta t by 2. So, if you look at this equation here what we did was we related s minus q delta t by 2 with s plus q delta t by 2 and minus q delta t. So, every step we had to find this q multiply by delta t and then subtract from s plus q delta t by 2. Sometimes we can directly prepare this plot similar to s plus q delta t by 2. We can prepare another plot which will have s minus delta q delta t by 2 and using this graph we can directly obtain the value without going through that additional step of computation. If we plot the inflow and outflow assuming that it returns back to its original position and as we said there will be a peak attenuation and lag. This point where the inflow and outflow intersect typically will have the largest outflow or the outflow will be maximum and the storage also. Similar to this q max we can plot a curve which shows t versus the level in the reservoir h. It will follow a curve similar to the outflow curve and it will reach a maximum since we say that q is proportional to h to the power 3 by 2. The h will be maximum wherever q is maximum. So, these two times will be the same if we say that q is proportional to h 3 by 2 which normally happens when we have un gated spillway. So, in that case the spillway would be like this that this charge through the spillway will be proportional to 3 by 2 power of h. So, the discharge will be maximum when the head is maximum. If we look at the equation we can see that d s by d t will be 0 when i is equal to o or if we use the symbol q then when inflow is equal to outflow then d s by d t will be equal to 0 and therefore, s will be maximum. Since we say that s is also related with h may be some power m may be some power m or may be a non-linear function of h s will be maximum at some point means that h will also be maximum at that point and q will also therefore, be maximum at that point. So, this shows that the intersection point where i is equal to o represents maximum outflow and maximum storage in the reservoir. In addition to this modified pulse method there is another method which is commonly used known as the Godrich method. In the Godrich method the same equations are written in slightly different form and the form which is used in this case is 2 s 2 by delta t plus q 2. So, it is almost same as the modified pulse method, but a little bit different in the coefficients and the way we modify the equation. So, now we can see that if we plot this term with h again we get a curve similar to this we can plot q also as we had done earlier, but here the computations become a little easier because now we do not have this delta q delta t term we have only this q term here. So, this 2 s 1 term is written as this term which is the same as this is obtained in a manner similar to that modified pulse method in which we find out the value find out h. Now, corresponding to this h we know the q and therefore, from this value which is known at the end of the time step we just subtract minus 2 q and we will get the value of 2 s over delta t minus q at the beginning of the next time step. So, this makes a little bit easier the computations of the water level under this charge at different times. The other method in addition to these graphical methods may be the use of a numerical method and these days with availability of computers, easy availability of computers and availability of various algorithms to numerically integrate a differential equation it is quite common to use numerical methods. For example, we can take the equation in its differential form the inflow mass outflow d s is the change in storage in time d t and what we can do is we can prepare graphs which relate the area of the reservoir. So, this is the plan area of the reservoir and we can write the change in storage for a little change in h. So, if we have a little change here d h for very small change in d h we can assume that the area is almost constant. So, d s will be given by a d h and this a will be a function of h. So, we can denote this a as a as function of h o is the outflow which we have already seen will be a function of h and i is the inflow which is a function of time. So, we get a differential equation in which the variables a are function of h outflow again a function of h the inflow is a function of time. So, the differential equation can be written as d h by d t because d s we have seen being equal to a d h. So, we can write d h by d t i as a function of t minus outflow which again let us denote by q and divide by the area which is a function of h. We can write this as some function of time and h where this is known function because area is a known function of head inflow is a known function of time and q is a known function of the head over the spell bay. So, this gives us a differential equation of the form d h by d t h and t or d n h which can be solved by using various available numerical techniques. One of the most popular numerical techniques is known as the fourth order Runge-Kota method and is normally taught in all numerical method courses. The equation which we use in this case is knowing the values at the ith interval we can write the values at the i plus first interval as a weighted mean of some d h by d t values which are denoted by k multiply by delta t and adding to the h at the previous time step. The values of k 1 k 2 etcetera are based on the function value h t. So, this is nothing but d h by d t denotes the slope of h versus t curve. So, d h by d t really indicates the slope at any point t. So, k 1 is also a slope, but evaluated at some point which is t i and h i. This is known to us because at t equal to t i we know h equal to h i. Our aim is to find out the value of h t i plus 1. So, what will be this value h i plus 1? First will be this value h i plus 1 whether the h curve will go like this or like this or like this. So, this is our aim finding out the value at the end of this time interval h i plus 1. So, what we do here since t i and h i are known we can find a slope d h by d t at the point t i h i. Using this slope we can go to a point which is midway between t i and t i plus 1. Let us call this point t i plus half and the value of h we assume at this point will be based on h i and then the slope which we have used or obtained this is based on the known value. So, k 1 is the value known at x i plus 1. So, this will give us a value of k 2 at half point corresponding to a depth which denotes the slope at the beginning moving up to the midway point. So, this is some h value which we predict based on the original slope of the line at h i t i and the time step delta t by 2. So, this k 1 is known from the initial conditions or at the beginning of the time step k 2 will be known at t i plus half and using the value of k 1 we predict a value at the midpoint. So, k 2 will also be known. Similarly, we can find out k 3 again at the midpoint, but using the new slope k 2. So, this delta t by 2 here and here they are because we are estimating the value at the midpoint and k 4 is evaluated at the end of the time interval. So, t i plus 1 h i, but now using a slope of k 3. So, k 3 delta t and now we use delta t because we want the value at the end of the time interval. So, what these k 1 k 2 k 3 k 4 represent are estimate of slope of the head versus time curve at different times k 1 at the time t i k 2 and k 3 at the time t i plus half and k 4 at t i plus 1 and using these slopes and giving them a weight 1 by 6 for k 1 1 by 3 for k 2 and k 3 and 1 by 6 for k 4. We get some kind of an average slope. So, we can see that this equation can be written as h i plus 1 equal to h i plus some average slope which we can call let us say d h by d t bar into delta t and this d h by d t is a weighted average of the values of d h by d t at beginning end and midpoints. So, this is nothing but 1 by 6 k 1 plus 1 by 3 k 2. So, we assign a weight of 1 by 6 to the first point and the last point and one third, one third to the two estimates which we have obtained for the midpoint. So, using this equation is starting from a known value of h i, we can estimate what will be h i plus 1 at the end of that time period delta t and then we can proceed once we know from a known value of t i and h i taking this delta t t i plus 1. So, using fourth order r k we estimate what is the value at t i plus 1 and then we proceed to the next step t i plus 2. Now, using these two as this is known and this is the unknown value. So, this numerical methods have become quite common recently because of the wide use of computers and easy availability of programs for numerical integration of differential equations. So, these are some methods which can be used to do the storage routing. Similar to this we have channel routing also in which lots of different methods are available. In the channel routing the basic methodology is knowing the inflow at certain point. We want to predict what will be the outflow at some other point and in this case therefore, as compared to the reservoir routing where we had a storage in the reservoir. Now, we have to think of the storage in the channel. So, storage in this area we can represent by let us say s. The channel storage can typically be divided into two parts. For example, if we have water level here and the water the bed level and the water level like this, then we could divide this into two parts. One part which has the same depth throughout and therefore, this will be in the shape of a prism. If you draw the cross section then this part is a prism of this section and of this length and therefore, the storage within this area where the depth is taken same as the downstream depth is called the prism storage. We typically can write this as s p for prism storage. The other portion which is above the prism and below the water surface up to the water level is in the shape of a wedge and is called wedge storage or s. So, total storage in the channel will be a sum of the prism storage and the wedge storage. This storage is used for the routing and the equation which we typically use is that prism storage will be a function of q because the prism storage is taken based on the downstream depth and the outflow the downstream depth will directly affect the outflow. Therefore, the prism storage will be a function of q. The wedge storage s w is typically a function of the nth. So, the total storage s can be written normally for channel routing. We assign some weight to the inflow some weight to the outflow and write the total storage as the wedge storage which is a function of i and prism storage which is a function of q. In this k is a coefficient which is sometimes called a storage time constant m would be an exponent which typically varies from about 0.6 for rectangular channels and it is about 1 for natural cross sections. So, m may vary from 0.6 to 1, x is a weight which decides how much weight we are giving to the inflow for the storage computations and how much weight to the outflow. S is the store in the channel. So, using this equation we can obtain various methods of routing the flood through the channel. So, knowing the inflow we again want to compute outflow q for the channel and one of the methods which is quite commonly used is known as the Muskingum method. In the Muskingum method we take m equal to 1. So, the equations which are written for q in terms of i and o are some function k x i plus 1 minus x. So, as we can compare with the previous equation this m value is taken as 1 in the Muskingum method. Therefore, we will have only i here and q here. x is the weight which is typically for the Muskingum method is taken to vary from 0 to 0.5. 0.5 means we are giving equal weight to the inflow and outflow and 0 means we are giving no weight to the inflow and weight of 1 to the outflow. So, if we put x equal to 0 then we get s equal to k q which is known as a linear reservoir and this is used sometimes in developing the instantaneous unidadographs which we have talked about already, but typically x value less than 0.5 will be used for Muskingum method about 0.3, 0.4 we can use. So, we can write this equation for the mass balance or the continuity using the Muskingum method. We can find the change in storage between two points which one before the interval and one after the end of the interval. So, with the time step of delta t s 2 is at the end of the time interval s 1 is in the beginning. Now, this s can be related with i and q using the equation Muskingum equation. So, m equal to 1 and therefore, we can get the change in storage by this equation. Now, this change s 2 minus s 1 from the continuity equation is also equal to i 2 plus i 1 by 2 delta t minus q 2 plus q 1 by 2 again multiplied by delta t. So, we get two equations for one for s 2 minus s 1 from the continuity, one from the Muskingum relationship here between s i and q and using these two equations equating them we can write q 2 which is the unknown in terms of the known quantities i 2, i 1 and q 1. So, i 2 and i 1 are known because i is known for all times q 1 is known because of the beginning of the time step we assume that q is known. This constant c 0 c 1 and c 2 are given as minus k x. So, c 0 is given by this c 1 is expressed as k x and c 2 is given as k 1 minus x minus delta t by 2 k 1 minus x plus delta t by 2. So, using these values of c 0 c 1 c 2 which can be obtained based on k n x and delta t we can write an equation which will be applied repeatedly for different time steps to obtain the unknown q 2 at the end of time step in terms of the known values of i 2, i 1 and q 1. Typically it is done in a tabular form to make the computations easier. There is sometimes it is also written as in a different form as q at any step i related with previous time step value some constant and delta in net inflow before the time step and net inflow after the time step. So, we can write an equation in this form also, but typically the Muskingham method using this form is. So, in today's lecture we have seen the flood routing. There are different methods of routing the flow either through a storage reservoir or through a channel. The storage reservoir routing is also called level pool routing because we assume that the reservoir level is horizontal and we do not account for water level variation inside the reservoir. And there are methods which can be used for the reservoir routing for example, the modified pulse method or got rich method and there are some methods like Muskingham method which can be used for channel routing. So, we have looked at the techniques used for different kinds of routing and in the next lecture we will look at some examples which will explain various aspects of designing for a flood computing the maximum flood discharge routing the floods through storage reservoirs and channels.