 Today, we shall be again continuing with our earlier topic, that is the gradually varied flow. Well, we were discussing about computation of gradually varied flow. And in that, under that topic we were discussing that different methods of computation. And we were talking about numerical methods also. and today we will be going to some extent in deeper of these numerical methods, not in detail but still we will be covering some of the aspect of that and then we shall be taking up some of the hands on some of the numerical problem that can help us in understanding the method we have discussed in a better way. Well with that thinking let us see see what we did in our last class that is we have discussed just to summarize till now what we have done we have discussed the classification of gradually varied flow that way we know that there are different classes then we did discuss about the characteristic of gradually varied flow different type of profile how they form and what are their characteristics at upstream what happened at downstream what happened and the where the profile is horizontal whether it is rising or whether it is falling that sort of things we have discussed and that very basic knowledge is important for computation and then we did discuss about the influence of control section that is if it if the profile is forming in a subcritical zone then we could see that control is at downstream if the profile is forming at supercritical zone then the control is at upstream so those aspect we did discuss and of course in the last class we again did discuss about the formation of profile over series of channels rather say channels which we can consider as series of channel of different slopes say slope is changing from milder to milder milder to steeper so that sort of things we did discuss and of course we while discussing those things we did not consider a slope change which can lead to another phenomenon called hydraulic jump because we are yet to study that one then after that we went for computation of gradually varied flow and in computation of gradually varied flow just to recall we studied particularly two method one is the direct steep method and the second one is the standard steep method well and of course there are some method like graphical method then direct integration method so this sort of methods are there which need recourse to graph and which need recourse to say tables and charts so as now in the days of computer we do not like to go for those things or rather we have other alternative way or say we can develop a computer program to solve this program very easily and that's why we are not using those methods and that's why we are not discussing those methods here here what we are discussing is standard steep method then direct steep method and of course we will be discussing numerical methods and in numerical methods as said earlier there are several numerical methods which can solve this sort of governing equation this sort of governing equation means the governing equation of gradually varied flow which is of the form that dy dx is equal to it is a function of x and y of course when it is a non prismatic channel then only it is becoming a function of x otherwise it is a function of y only but it is not a function of x only and that's why we cannot solve it by analytical procedure and because this is a non-linear differential equation so this non-linear differential equation for solution of that we need to take recourse to different methods and these are say one is direct step method one is our standard step method then we need to discuss about numerical methods okay so let us now discuss on these numerical methods of course we have discussed on direct step method already and standard step method and we will be taking up some numerical problem for solving this standard step method here just to have a field of the actual problem well well so for numerical methods are concerned in fact there are several numerical methods that we did discuss in the last class for solving this sort of equation that is dy dx is equal to rather in not writing the directly governing equation visually very flow if it is f x y and for solving this sort of equation we have different type of numerical methods and here we are discussing only a few of those methods and we are not discussing more details of those methods but just to understand how basically the concept of numerical applications are applied we will be explaining that with the help of some graphical representation well so we are taking up the first one this is Euler's method say what basically the numerical methods do to see that suppose we have a curve it may be gradually very flow it may be anything well now this curve has a equation like this dy dx is equal to f of x y it is a function of x y in our case this equation is dy dx is equal to say sb minus sf divided by 1 minus q square d by g a q well in case of gradually very flow otherwise it can be any equation well now if we just know the initial value suppose in this direction we are having y in this direction we are having x well if we know the initial value at x equal to 0 then say this is equal to y 0 suppose at x equal to 0 we know the initial value of the y then for this sort of situation which we can call that initial value problem we know this initial value and then we need to solve for the other value then this method is quite advantageous how say when we know the initial value y 0 and we have an expression and of course this is at x 0 that also we know and we have an expression for the slope dy dx so if you put this value y 0 and x equal to say x 0 and not write that 0 at x equal to x 0 this can be any value also it can have any value not necessarily 0 okay so at equal to x 0 so we know this value so if we put this y 0 and x 0 then we get the dy dx at x equal to 0 and x equal to x 0 and y equal to y 0 so that way this value is known what is dy dx it is nothing but the slope at this point slope at this point if I draw the tangent to this point then of course my line is becoming almost horizontal because there should be some slope in this let me take the curve like this so say at this point if I draw a slope it is going like that and this slope is nothing but a tan of this slope rather is called dy dx then theta we can write okay now in numerical method what is that we know this value and then if we take a small distance in x direction if we take a small distance in x direction okay I am taking it here I am taking it here say this is a very small distance delta x okay or in many numerical method it is written as h a small distance h okay now so what about this distance as we know this slope dy dx and if we write this distance as delta y and this is say delta x then we can write that tan theta is nothing but delta y by delta x or we can write the dy dx at say point y is equal to y 0 x equal to x 0 this is equal to say delta y by delta x well so from this we can find out what is delta y so delta y is nothing but delta x into say dy dx at say point x 0 y 0 well so by the simple expression we can find out delta y and this is a horizontal line we are drawing so up to this much distance this distance is nothing but y 0 well now the distance y 0 plus delta y will give us up to this particular point so now if I just draw it in an enlarge form then I can show that y 0 which is the point actually our interest we are interested to know that at x equal to x 1 means x equal to x 1 means say delta x is what delta x is nothing but x 1 minus x 0 at x equal to x 1 what is the y so our actual value for y 1 is this much is this much our actual value for y 1 is this much as per this curve but if this delta x is very small then we can write that delta x is equal to y 0 plus delta y approximately equal to we can call now of course from this curve it is not very it is quite visible that y 1 and y 0 plus delta y these are not equal but if I consider my delta x to be very small now you can see suppose my delta x is here my delta x is here this is my very small delta x small delta x if my delta x is only this much then with this being our delta x then say this will be small distance will be delta y this small distance will be small delta y say and then you can see these two points that is the point of the curve and point on the tangent are almost at the same point well so this delta x being very small we can very well write that the delta x sorry this is not delta x we should have written it y 1 y 1 is equal to y 0 plus delta y that we can write very well now so this is our y 1 and we can write that y 1 is equal to y 0 plus this small delta this is valid for a small value of delta x but if we increase our delta x to larger value if we increase our delta x to larger value then this is introducing some error now once we get the y 1 once we get the y 1 then this y become y 1 become our known value then we can take a slope at this point suppose y 1 is known well let me just take it if y 1 is known then we can find y 2 is equal to when y 1 is known you can find out the slope say again delta x say next delta x you are taking delta x into say dy dx at y 1 x 1 so this value dy dx you can calculate at y 1 x 1 as now you know x 1 value and y 1 value knowing the y 1 and x 1 you can calculate dy dx then multiply this by delta x and you can get your next y 2 that way you can continue solving for values of depth y for different x but from this graphical diagram or from this graphical representation one point is very clear that if our delta x is not very small then if our delta x is not very small then we can end up in getting suppose our actual curve is like this and with our numerical value we can have suppose this point if we consider is this point then this point we can get suppose here then you are calculating based on this y 1 value say for this x 1 x 2 x 3 like that we are calculating and for this value if you calculate the slope again say you are getting this value first then at this point you are taking slope and then you are getting another value here and at this point you are taking slope and then you are getting another value here so your curve will be like this it is basically some discrete point you are getting and your actual curve is this one and but your computed curve will be this one this will be your computed and this will be your actual so that way you can always end up introducing some amount of error if the curvature is in the downward side the things will be in the opposite direction it will be away towards upward direction so that is though it is very simple Euler's method but it can have some amount of error provided your delta x is not very small again not very small is quite blurred that what we mean by very small that is also we need to see into that particular aspect well then because of this particular drawback this was improved based on some of the consideration that that is called as improved Euler method well say this is our graph and now we can see say if with dy dx at this point we are getting slope at this point and then we are suppose our delta x is this one i am drawing a large delta x and then for this delta x say my delta y initially calculated will be this one delta y and this being y 0 this is y 0 and from here to here it will be y 1 that is y 1 is equal to y 0 plus delta y well now we can see that if we take again with this known value y 1 if we take a slope if we take a slope this slope at this point or rather for this y 1 value the slope will be like this slope will be like this now based on this slope based on this slope if we calculate suppose this slope we are considering that this slope is not there rather here then if we calculate let me draw the slope here if we calculate here then our let me take the slope like this if i draw the slope here then it will be going this part so based on this slope slope at the point y 1 and y y 1 if we calculate our delta y then delta y will be large like this this is say delta y we are calculating based on the slope at this point but slope we are using the slope at this point but putting the slope at initial point y 0 okay so this time our y 1 will be quite large like this and now if we take say this is our y 1 computed for the first time and this is say y 1 computed for the second time and this is the delta 1 computed for the second time this is delta y 1 computed for the first time the second time means with the on the basis of the slope taking at the value of y 1 which we have computed okay now if we do like that then we are getting the value in different way now if we take average of these two value that is average of y 1 computed with this level and y 1 computed with the first one then if we take average the value will be coming somewhere here somewhere here and you can see very well that is average of this means your depth is neither this one and this one you are getting average this is your y 1 and that average y 1 is you can write this average y 1 is equal to say y 0 plus delta y 1 plus delta y 1 this is y 1 1 okay let me write little larger here say delta y 1 1 plus delta y 1 2 divided by 2 well so that way we are getting this expression and what is this delta y 1 that is nothing but we have already explained what that delta y 1 that is nothing but delta x into dy dx so what we can write here say y 1 is equal to y 0 plus delta x into say or you can write delta x by 2 or you can write half of dy dx at say 0.0 and plus dy dx at 0.1 well so this way we can compute the y 1 value and this y 1 value will be more closer than the earlier one more closer than the earlier one well so this method is referred as improved Euler method that means we are using the slope at the initial point and then we compute the second point and based on the slope at the second point based on the known value of y 1 in the second point we again compute the slope and then we take average of these two slope and then we calculate this y 1 and if the variation of this slope if the variation of this slope see slope is changing here it is this must here it is this must so slope is changing now if the change of slope between this point and that point is linear linearly if the slope is changing then this method will give accurate result but of course the the curves are not like that that the slope will be changing linearly and it may not be this for all situation and that's why it will not give you that accurate result and of course this method after this method this has another advantage now so what we have seen we are taking average of this slope average of this two slope and so once you we calculate this y 1 and say once we calculate this y 0 okay y 0 and then we have calculated the y 1 and then we want to calculate y 2 y 2 rather than again doing like that taking a slope here and then taking a slope at this point taking a slope at this point and taking half rather than doing all those things what we can do that for computing this value for computing this value we can see that if we use the slope at this point if we use the slope at this point that is at the middle point and then if we calculate from the beginning of this one using the slope at the middle point we are reaching much nearer to the value of the second point if we consider this as 0 then this is becoming the two second means that it is actually the third point nomenclosure is becoming as y 2 so what we can do y 2 is equal to we can write y 0 plus we are using this time our delta x is 2 delta x twice of delta x into slope we are using dy dx at say y 1 so to calculate the value of y 2 we are using y 0 and twice delta x then dy dx at y 1 and in general if we want to write this in general we can write that y say if this is i this is i plus 1 we are interested to find i plus 1 i minus 1 is also known to us so what we can do that for calculating i plus 1 we can write y i plus 1 is equal to y i minus 1 plus 2 into delta x into we can write this as y dash y dash means y dash represents a dy dx at i equal to 1 sorry at i mean dy dx at the point i middle point so this procedure can be adopted this is basically the concept is similar to this one rather than taking average we are using the slope directly of the center point for that first we must find this initial point and once we find this initial point then from using this slope at this point we can and starting from this point we can find the third point once we find this point then starting from this point starting from the previous point and using the slope at the second point we can get the the third point so like using this way that improve Euler method can be used well but when we have computer we can go by this method also from one point to other point well then it was found that this method as I was telling that if this is quite if this point is say linear then of course we can get a slope which can directly lead us to this point but because of that limitation and again this Euler method was modified and this modified Euler method was used by process with some modification of course and then he applied it for gradually very competition well and what is that particular method is that say we are trying to compute this based on this y 0 we are trying to compute the y 1 and the y 1 that we are getting this is delta 1 and that suppose I am now numbering as delta 1 1 that is the first time what we are calculating and then what value we are getting that we are numbering at y 1 then on the top we are writing 1 so that means for the first time based on this slope when based on this known value of y 0 we are calculating slope that is dy dx and then we are getting a value y 1 that is our say first predicted value of y 1 then we calculate the slope at this point we calculate the slope at this point because y 1 is known and that slope we are putting here and we are getting another value of y 1 another value of y 1 this y 1 based on this y 1 we are again calculating this is leading us to a slope y 1 dash so finally we are calculating taking the average slope of this one and that one we are drawing a line and we are calculating a value of y 1 just like the improve Euler's method so this value we are getting that means what we are getting say y 1 y 1 our first trial y 1 is equal to y 0 plus y 0 plus delta x into y 0 dash plus y 1 dash this dash means well the symbol dash y 0 dash or y 1 dash or any value y i dash is equal to say d y d x at i that is what we are meaning in this one represent that this is the first predicted value d y d x and this plus this divided by 2 ok this plus this divided by 2 then we calculate using this y 1 this we are not considering as the correct value in improved Euler method we consider that this is the correct value but here we are not considering this as the correct value using this y 1 now again we will be calculating another second trial second trial that is we are considering iteration but for the same value of y 1 what we are doing that y 0 plus delta x by 2 and then we are trying y 0 dash plus y 1 1 dash that means whatever we considered as y 1 in the previous step that we are using here as the slope and here we are using that y 1 and we are calculating the slope and then we are trying a next value or next trial value of y 1 now if this value and that value is almost equal then we can consider that this is the correct value but if it is not equal then we will go for another trial for the same y 1 value but we are doing for the same y 1 value we will go for another trial that again y 0 plus delta x by 2 into y naught dash plus y 1 2 dash with this we are calculating a slope with this y 1 second trial value we are calculating the slope at that point then and we are taking half of that and then we are using this delta x and we are trying to find that so this procedure will be repeated till this y 1 2 and y 1 3 that is 2 consecutive depth calculated by this procedure become almost equal and once we get that this is almost equal then we know that our calculation is correct or and we can consider this to be correct value of course how much accurately we can approach this value that will again depends on whether this this will lead us to almost correct value when our variation of slope is linear if it is not linear again there will be some amount of error introduced here well this is what the modified Euler's method and what Prasad did rather than comparing this y 1 value that is the consecutive depth he compared the consecutive slope that is being computed in fact it is indirectly the same thing if you compare the slope if it is approaching the correct value then also you can have this correct value so that method was used but in all this method we have a drawback that if we increase the delta x then we cannot reach the correct value because the profile equation is of the type that it will not lead you to the correct value in although you carry out all those iteration so we attempted this try to rectify this particular drawback and what was done this is also basically a corrected initial value is calculated then iteration is carried out but with the concept that this say this is the curve and suppose we have this these are our delta x value these are our delta x value then for computing from say first delta x this way it is going there and for computing from this known value y 0 for this y 1 in earlier method also in Prasad's method also we do iteration but here what we are doing say if this is the slope we are arriving at this point then this method can be applied for only gradually varied flow the reason is that the dy dx the slope is a function of y only dy dx is the slope this slope is a function of y only so once we know the y then we can calculate dy dx but if it would have been a function of y and x then this method cannot be applied because there we will have to know the x value also here what the advantage is that once we know the y 1 calculated then this y 1 actually is the depth at this point and so once we know this y 1 then we can find out what is the corresponding value of this x what is the corresponding value of this say small x delta x what is the corresponding value of this delta x for this y 1 how we can do that because that is very simply this is our delta y this is our delta y and this enter depth we are first initial trial y 1 dash and then very basic concept is if our this length is small now we are coming to this point if we solve at this point the suppose this is y 1 so slope we are getting at this point and if this portion is in the small portion if we consider the small portion if we can consider to have linear variation of slope then average of slope at this point and slope at this point will pass through this point obviously and therefore based on this point we can find out what is our delta y value because delta y is this one obviously so delta y by this small delta x is equal to actually slope this one which is average of the slope at this point and slope corresponding to this y 1 so from that we can find out this delta x if we mark this as delta x 1 then what we can do when in the first trial we were using our entire length as delta x in the second trial we will just deduct this x 1 and our only remaining length for this first delta x first del x will be only this much so considering this as our delta x now say delta x 1 this as delta x 1 what we will do again we will start from this point and we will go to this particular point and then we will again come back and then we assume that within this small portion this variation of linear slope is quite valid and then we can calculate this small delta x 2 so this way we are reducing the length or extent that we are using for computing the incremental length so as we are reducing our delta x in each of the repetition or each of the iteration calculation of delta y for this small delta x is becoming more correct and the assumption of linear variation of slope between the small delta x portion is more valid rather than considering a large delta x that is why even if we take a large delta x say from here to here it is significantly large but when we calculate this value we are transferring this value to this point and our small delta x are getting reduced this value are getting reduced and we are successfully using smaller delta x and so this method lead us to and finally when we see that our no more small delta x is remaining no more that is we cannot come back that is no more small delta x is remaining then and we can see that actually our delta x should be equal to summation of all delta x 1 then we are always checking after each of the calculation we are checking that if this delta x summation of delta x is becoming equal to the delta x if not we are continuing and when the summation of delta x is becoming almost equal to the delta x we can stop that means our entire delta x portion has been computed so what value we are getting that is our final value of y 1 dash y 1 final value of y 1 so again from this y 1 value to compute the y 2 value will be proceeding in the same way dividing it into dividing means we are not dividing rather it is automatically getting divided and finally we are checking if this delta x is complete then we will get this value and then this method when you compare with now Prasad method and other methods say Euler methods and Ranga Kutta method also we compared then we found that this method is giving better result in case of gradually varied flow computation of course we must know that this is not a generalized equation which can be a generalized method which can be used for solving any sort of I mean any sort of differential equation of this type it can be used only when this is a function of y when this dy dx is a function of y and as the gradually varied flow is having that advantage when we consider one-dimensional prismatic channel then we can very well apply this method well with this introduction to numerical method and of course as I was telling that there are methods like 4th order Ranga Kutta method is also very accurate and it keeps correct result and then let us take up some problem of course numerical problem I mean this numerical methods we cannot apply here because it will be quite iterative and it will require lots of time so let us see and let us have one problem which we can solve in this class and we can have a field of actual gradually varied flow profile computation well so we can concentrate into the slide say a rectangular channel of 6 meter width so channel is 6 meter and of bed slope 6 into 10 to the power minus 3 is carrying a discharge of 5 meter cube per second then Manning's coefficient of course Manning's coefficient we need to assume Manning's coefficient can be considered as say 0.014 well that is the n value and what was done if you remember this particular problem we took when we were discussing computation of uniform flow and for our purpose at the time we were computing what is the uniform flow date and that is why so for the first part of the problem is concerned I am taking the same problem so that computation of uniform flow part we can avoid or rather we can simply just give the value as this was already done in class of computation of uniform flow well so that means up to computation of y n we know well then what has happened a barrier of height 1 meter was constructed across the channel so many a time in the channel for various purpose say for irrigation purpose sometimes people give a barrier at the upstream downstream so that water level can be raised many a time for fishing purpose also people give barrier in the interior area particularly in Indian condition I am talking about in Bangladesh also even and then people give the barrier to obstruct the flow and then they can get sufficient amount of water they can go for agriculture somewhere for fishing purpose also people give obstruction that way some purpose for many different reason we can give obstruction so 1 meter of obstruction was constructed it is small obstruction across the channel however small means considering the other values this may be quite significant sometimes so this 1 meter obstruction was constructed across the channel to raise the water level to raise the water level so objective was that we will raise the water level on upstream now when we are raising the water level that means definitely there is a change in the water level and when there is a change in the water level the profile will be formed of course the change can lead to another phenomenon called hydraulic jump that will be coming in the next class but gradually very flow profile will also be there well now we are of course concerned about the gradually very flow component so compute the flow profile on upstream of the barrier so what sort of flow profile it will be that is what the first part of the problem and then there is a second part if the slope of the channel is reduced if the slope is reduced then what sort of profile will be forming on the upstream many a time when we give obstruction then if the water is carrying lot of sediment then as the flow velocity get reduced its sediment carrying capacity also get reduced and all the sediments may get deposited into the bed and when the sediments start depositing into the bed bed slope may change that is naturally of course but for some of our own purpose sometimes we may reduce the slope ourselves by say putting some soil there or just we can augment the bed slope to some extent so that is whatever may be the reason or whatever may be the process but suppose bed slope got reduced to 1 into 10 to the power minus 4 what sort of profile will be forming on upstream then how the profile is changing well then another question is being added another part is there if height of the barrier is increased now suppose height of the barrier is increased we were originally putting a barrier of 1 meter now due to some reason to increase the depth of profile we thought that okay we will be increasing the depth of you will be increasing the height of the barrier so that we can raise the water level further then and of course slope suppose whatever we did change that is rare there I mean change slope is there and what change will occur in the flow profile and ultimately our target is there we need to first know what type of profile will be forming in different situation and then we need to compute the profile for this situation this situation means of course taking up all the situations may not be possible so we will just see for the last situation how to compute the profile well so let us go by one by one well so this is our say well I will have to take the arrow this is the bed originally which had a slope as it was explained which had a slope of 6 into 10 to the power minus 3 okay now in this slope we had that this sort of uniform flow is occurring so we need to know the depth of uniform flow and that we have already done in our computation of uniform flow okay now here we are putting a barrier then this uniform flow will not be there rather a gradually varied flow will be coming like that and then we need to compute this gradually varied flow profile okay well now what we can see that this is the bed slope and this was the original profile this was the original profile original uniform flow depth and as per our computation of uniform flow depth we had our yn that is the normal depth is equal to this normal depth was equal to 0.3 0.335 yes it was 0.335 yn I am not computing that right now you can use the approximate formula of dash and bar and then you can carry out some iteration to see whether it is satisfying that required discharge or not and this discharge equal to 5 q max and then we can have these things then once we know the yn initially in fact we are not knowing what about this profile whether this profile is like that then we need to calculate what is yc critical depth well critical depth as we know it is equal to q square by g and of course in the problem we are considering a rectangular channel q square by g whole to the power one third well so this value we can have q square is nothing but or 5 is the q max total discharge 6 is the bed width so this is the unit discharge and then 9.81 and whole to the power one by three so that way we can have the yc value as equal to 0.414 0.414 meter well that means the value is critical depth value if I draw here this will be this will be say 0.414 yc then the height of the barrier is given suppose one meter height of the barrier is one meter now whether our profile will be of this type or it will be of different type that we need to see definitely this is initially with a drone from our general concept that water will be rising but how the water water will be rising whether this will be this one or it will be something else that we need to see so when water need to cross this level water need to cross this level suppose this one meter height of course water will not be crossing just at one meter there will have to be some depth here which we can find suppose width of the channel is known width of the channel is known say six meter so height we can find out using the formula of where that is two by three that is this y also we can find out that two by three cd two by three cdb root over twice g h to the power three by two using this relation we can find out what that height is q is equal to this one so our q is known we know the cd we will have to assume this is nearly equal to 0.62 for rectangular section and then b is known g is known it is twice g then h we need to find out so from this formula we can definitely find out this h value however this h value will be will not be that high as compared to this one meter that's why for many practical purpose we can neglect this depth also and suppose when depth of barrier is one meter we can consider the flow to be also of one meter flow depth to be also of one meter but to be more accurate we need to calculate this and for some of the combination this h this h may be quite significant and as such we can check that amount also right now in this problem we are not doing this we are considering that depth is one meter means or barrier height is one meter means at that point depth will be one meter well now the problem is that initially the water is coming as normal depth uniform flow at the depth of point three three five meter finally it is moving at the depth of say one meter finally it is moving at the depth of one meter okay and from this we can see that critical depth is point four one one four one four so this part of flow initially it was supercritical flow initially it was supercritical flow and after giving the barrier it is rising to a subcritical flow so our profile will be here in the subcritical flow portion and normally when flow is occurring y c is greater than y n that means it is a steep slope that means it is a steep slope the profile over steep slope and this is in zone one so it is basically s one profile and when it is s one profile when it is s one profile then what we can do that this is our known depth and we can start computing from this part it is a rising profile and we can start computing from this part and of course although initially we thought that the profile will be like this actually it will not be like this because profile meet the normal depth line asymptotically not the critical depth line it will meet critical depth line normally so the profile is coming like this so when we want to compute that profile we will have to start from this particular depth one meter plus something of course then we need to come down by direct step method or numerical method whatever it is we can come down and we can compute it up to a certain point up to critical depth point however in reality it will not come to the critical depth point the reason is that here it is supercritical flow and from supercritical to subcritical when the profile moves then there is lot of turbulence and it will be forming a another flow phenomenon that is called hydraulic jump that will be discussing just after discussed so right now we are not talking about this part but this will be a hydraulic jump and that part is not changing because disturbance is going on at the downstream side and this is a supercritical flow this is a supercritical flow so this flow is not getting affected that means this flow surface is uniform flow it is coming as uniform flow and from this particular point it is jumping like that and of course for some other situation when this point will be coming here and then there may be some other influence on this that will be coming when we will be discussing hydraulic jump and then we will be discussing more about that well right now our point is that this is s1 profile and we can compute this profile in this way now if we go to the second point if we go to our second question if we go to our second question that we are getting say if the slope of the channel is reduced to 1 into 10 to the power 4 well if we reduce the slope like this the slope we are changing to 1 into 10 to the power minus 4 then if we compute the normal depth y n our critical depth is remaining as it is and critical depth is it is 0.414 and normal depth will increase and it is found that normal depth goes up to depth of 1 point y n computation of normal depth i am not doing here but y n is equal to 1.268 meter so that we are getting and our barrier height now is 1 meter barrier height is 1 meter so what is that when barrier height is 1 meter and uniform flow on the downstream side whatever way it cross these things say uniform flow depth will be this mass and then our entire suppose it is crossing this part and then there will be some obstruction and then it is moving again ultimately it will be moving in this part and of course suppose if we assume that this part slope is remaining as it is as like the earlier one so what it was that is it is earlier slope was earlier slope was this slope is 6 into 10 to the power minus 3 suppose we consider that after giving barrier we are changing these things then it is 6 into 10 to the power minus 3 this is suppose slope in the downstream this is remaining as the original one then our normal depth here will be again if we compute this will be the y and to what we did compute earlier this is equal to say 0.335 so from here flow will have to come down like this and it will be falling here and then it is meeting this point and ultimately it is going like this so this part of flow it is in mild slope now it is in mild slope because your critical depth is this one and normal depth is this one so normal depth being greater than critical depth it is mild slope and it is in zone 2 this zone is zone 2 so this profile is m2 and the depth here will be this much so we can start computing from this point and again we can come back and we can compute up to the point where it is becoming almost equal to yn now let me take the third case in the third case it is said that the barrier is increased to barrier height is increased to say 2 meter barrier height is increased to 2 meter and the other things remaining same our critical depth here will be critical depth here will be say 0.414 and normal depth is now say if I draw the normal depth line it will be 1.268 and barrier height is 2 meter so as the barrier height is 2 meter the profile here will be rising like that and then it is crossing this part crossing this part okay and so this profile is as yn is greater than yc yn is greater than yc so this is mild slope and this is zone 1 because it is above this one so it will be m1 profile now for computing this profile we need to start from this one say again it will also be greater than 2 by the small amount but if we start from 2 then we need to compute up to just little higher than this yn okay now how we can achieve this by standard step method let us see briefly say y1 we why first we are starting from the depth 2 well then for 2 as the depth is known width is known we can calculate the area this is being calculated 6 into 2 12 then perimeter we can calculate then hydraulic radius a by p we can calculate then velocity we can calculate q by a q is equal to 0.5 that is known so we can calculate this one then energy we can calculate y plus v square by twice g so this we can calculate that so up to this much suppose we are taking that at 2 this is the point at 1.8 next depth suppose we are considering as 1.8 this is at 2 it is at 2 and yn is 1.268 so we are starting from 2 and then at this point we have calculated everything then we take let us take another depth 1.8 then 1.6 will be next 1.4 will be next then this just higher than the last depth it will be say 1.27 little more than 2 1.268 little more than 1.268 we are taking 1.27 which is just higher than yn so for 2 we are getting everything then for 1.8 up to this much we can get now knowing these things and then energy delta e because our delta x is equal to this delta x we can calculate this delta x is equal to delta e divided by sb minus sf so this delta e we can calculate by this minus that will give us this one this minus that we are giving this one so at every section for each section we are calculating the e and delta e then friction slope by using manning's formula we can calculate for each of the depth then average friction slope we can calculate as this plus this this plus this divided by 2 so that way average friction slope we are getting and then we can find out this delta x by using this relation that is delta e is equal to delta x is equal to delta e divided by sb minus sf this divided by sb minus sf so we are getting these are the delta x so delta x 1 delta x 2 i am writing dx there dx 1 dx 2 like that dx 3 and then you can see for this small rise of 2 meter depth in a 6 meter wide channel what is our extent total length of the gradually very profile entire length is becoming 20553 that means 20 kilometer 205 and then for a first part as it is rising any at this point it is rising so for lowering of 0.2 meter extent of slope is 2.8 kilometer or 2881 meter but here as we are coming here it is meeting asymptotically means for the same drop here also the drop is 0.2 meter from 1.6 to 1.4 but in this part in this part the extent is length is delta x is quite large so actually though i am drawing it like that the profile will not be of this sort it can be it will be of the type that for equal amount of drop for equal amount of drop this is gradually increasing for equal amount of drop drop is may be 0.2 everywhere this drop is 0.2 but this dx is increasing dx 1 dx 2 dx 3 like that it is increasing and finally for all if we sum up all these things we are getting total length of gradually very profile well we have just tried our level best too so how a problem can be solved it is definitely difficult for taking up solution of problem in this sort of class but still we have tried our level best and in the next class we will be moving to rapidly varied flow well there we will see how we can compute for rapidly varied flow and how with gradually varied flow and rapidly varied flow being in the same situation we can have combination of this sort of flow so thank you very much