 Welcome back to our lecture series on advanced hydraulics. So, for last few weeks, we are dealing with the module on uniform flows and in the last class, we have started the portion on uniform flow computations. We had elaborated various methods on that. We were discussing on the various methodologies of for computing uniform flow and raw. First let us start with the quiz for the last class. In the last class, whatever we have taught based on that, we will take a brief quiz. After that, we will just go through that briefly again, those portions and today we will start the next portion of the module. So, please begin your quiz. You will be given 5 minutes to answer the quiz. What question for the quiz is, what is Manning's formula? So, you please write down what is Manning's formula. So, very easy one. The second question for you is, enumerate 5 factors that may affect the Manning's roughness coefficient. You know the Manning's formula. In that there is a coefficient called Manning's roughness coefficient. So, what are the 5 factors? Just enumerate some 5 factors. There are many factors that affect Manning's roughness coefficient. You enumerate any 5 factors that may affect the Manning's roughness coefficient. I hope you have answered that. The next one is, what is the depth of water in uniform flow call? What is the depth of water in uniform flow call? It is just a one word answer. These answers should be given in one or two words. Now, the last question for you is, what is convinced factor for uniform flow? What is convinced factor for uniform flow? I hope you have answered that. So, this quiz was just a brief quiz just to test your updation of the course, how you are going in proper time and proper methodology. So, we hope that you have answered all the questions. The solutions for this quiz is Manning's formula. I will just write it. So, the solutions first one, the average velocity in any cross section of the channel for uniform flow. It is nothing but 1 by n r to the power of 2 by 3 s naught to the power of half where n is called Manning's coefficient. Roughness coefficient r is your hydraulic radius s naught is your bed slope. The solution for your second question, your second question was enumerate 5 factors that may affect Manning's roughness coefficient. You can enumerate any one of them. You can tell that say surface roughness for example, you can tell channel irregularity, you can also tell channel alignment, silting, scouring. Of course, one important point, vegetation. Vegetation also considerably affects your roughness coefficient. If any obstruction is there in the channel, seasonal changes. So, you can enumerate any one of any 5 of this or even if you know more than this thing that can also be that is also most welcome from your side. So, what was your third question? The third question is what is the depth of water in uniform flow called. I will just mention it here itself. It is called normal depth. No need to further elaborate the expression. Then the next question is what is conveyance factor for uniform flow. If you recall the discharge, the uniform flow discharge it was given as a function of normal depth y n and Manning's roughness this thing into s naught to the power of half. This function of normal depth y n and Manning's roughness coefficient n was subsequently given as conveyance factor k into s naught to the power of half. So, your conveyance factor k is a function of your conveyance factor k is a function of normal depth y n and roughness coefficient n. It is given as 1 by n a r to the power of 2 by 3. Thank you. Then let us start today's lecture now. So, today we will be continuing with uniform flow computations. So, in the last class we had studied on Manning's equation. We have studied normal depth conveyance factor section factor. We have also briefly discussed that what are the procedures to procedures to compute normal depth. So, in that we mentioned that you can use design charts, you can employ trial and error method, you can use numerical methods. So, how to develop your design charts? This was explained in the last class. We had in fact developed some design charts for trapezoidal channel or how the table how to compute design charts and all that we have done it. We have derived it in fact derived it for trapezoidal channel for any side slope 1 is to b. If you recall them any slope 1 is to b and b is equal to 1, when we substituted that the corresponding table was also developed. You can see how the trial and error method is also employed to compute normal depth and how to use numerical methods also to compute normal depth. We will just see it in the following portion now. So, the trial and error method. So, you recall the Manning's equation for discharge 1 by n r to the power of 2 by 3 s naught to the power of half or your discharge q, this is 1 by n a r to the power of 2 by 3 s naught to the power of half. In this relationship you if you remember them quite you had obtained the following relationship a r to the power of 2 by 3, this was club together this is nothing but n times the discharge q by s naught to the power of half. You had obtained such a relationship in the last class, subsequently we suggested this quantity as a section factor and all if you recall them. Now, the point here is to use your trial and error method in the left hand side, this is your left hand side, this is your right hand side for a given channel usually n is a given quantity q is also specified for you, bed slope is also given or it is a channel property. You are now requested to find the normal depth y n that is the objective here that is you are using the trial and error method to compute your normal depth. So, how will you compute normal depth using trial and error procedure? Again just as an illustration let us go with the trapezoidal channel, we are following the same norms as we are doing in our earlier classes. The depth of uniform flow is called normal depth, the bottom width is b 0, side slope 1 is to b. If you recall the area of cross section for such a trapezoidal channel is y n into b 0 plus small b into y n, your wetted perimeter p is equal to b 0 plus 2 y n into root of 1 plus b square. Therefore, your hydraulic radius r is equal to a by p y n by b 0 plus 2 y n into root of 1 plus b square. Therefore, in your relationship a r to the power of 2 by 3 is equal to n q by s naught to the power of half. Substitute the quantities whatever relationship here with respect to y n you substitute it here. Now, what about the r h s term? Your r h s term here all the quantities are known all the quantities here are known to you. So, you can just specify it as a known quantity j. So, you are now going to do your equating a r to the power of 2 by 3 is equal to a known quantity j. Further proceed substitute the value of a and r you will see that you are getting the following relationship y n into b 0 plus b y n square b 0 plus 2 y n into root of 1 plus b square whole to the power of 2 by 3. This is equal to a known quantity j. You can do one thing you can just bring all the terms here. So, b 0 plus b into y n whole to the power of 5 by 3 into y n whole to the power of 5 by 3 by b 0 plus 2 y y n root of 1 plus b square whole to the power of 2 by 3 this is equal to a known quantity j. So, once you have such a relationship now you are having such a relationship. Now, substitute for various value of y you try with various values of y n. You see which value of which value of y n satisfy your equation one satisfies equation one you have to find that which value of y n satisfy that equation one you have to select that this is called the trial and error method. So, you have to try it for several values of y may be some of you if you are lucky you may get it in 2 or 3 trials itself. Some of you may have to go for some nearly 100 trials then only they may arrive at the solution depends on luck also and also through intuition based on your problem and all what could be the initial that means or guess value how you try initial guess and all how you do that depends on your intuition as well. So, you can try it many problems for various this is only for a trapezoidal cross section. You can try it for various. So, similar to similar to equation one for other cross sections you can develop corresponding equations. Now, it is up to you say whether it is for rectangular triangular even circular it is up to you to do work it out and just find it what are those corresponding equation try the trial and I mean use the trial and error method to find the normal depth. Now, let us go with the next method. So, now next method we are going to do is the how to apply numerical methods to obtain the normal depth for uniform flow. Here if you recall equation one previously developed that is B 0 plus B y n whole to the power of 5 by 3 y n whole to the power of 5 by 3 by B 0 plus 2 y n 1 plus B square this equation it was of course, developed for the trapezoidal channel. You can obtain similar equations for other cross sectional channels also in this equation this is a non-linear equation in y n. So, to solve this equation you have to use non-linear solution techniques using the numerical methods there are methods or in the new methods or techniques to solve non-linear equations. So, you have to employ them here to solve your equation and get the normal depth some of them I will just briefly mention that you might have heard about Newton Raphson method conjugate method biconjugate method all secant method there are various methods to solve non-linear equations in the numerical methods portion. So, I am not going to describe all of them just briefly we will see through the Newton Newton Raphson method. So, your equation I can now obtain as equation from this thing B 0 B y n again I am iterating here iterating that this equation was developed for the trapezoidal channel for other cross section it is quite different. So, in this equation LHS is unknown RHS is a known quantity fine. So, the Newton Raphson iteration or Newton Raphson method it suggests that if you can develop a function of y n and if that function is differentiable. So, you will you can obtain the first differentiation f dash y n you can iterate and obtain the value of y as y n nu is equal to y n old minus f of y n by f dash y n this is the standard Newton Raphson iteration or Newton Raphson method to find non-linear solution. So, here what does y n nu and y n old mean this previous value or guessed value y n nu means it is the modified value or improved value. So, the crux of that is that if you have any non-linear equation in y n you can give an initial guess of y n some value and start using this relationship you first obtain a function of y n which is also possible to be differentiated then using your initial guess or old guess you can get a new value of y n or improved value of y n using the given equation mentioned here. So, that we can employ here in our case for the open channel for a computation of uniform flow we can suggest that your f of y n is equal to b 0 plus b y n whole to the power of 5 by 3 y n to the power of 5 by 3 minus j. Let us consider that your function of y n is equal to the following is in the following form. In this following form now if you obtain this function this is also quite possible to be differentiated this was for trapezoidal for a general case you can suggest f of y n for any cross sectional method you can suggest f of y n is equal to a r to the power of 2 by 3 minus j for exact value of y n this relationship a r 2 to the power of 3 minus j should be equal to 0. Now your initial guess y n initial it may not equal to be your y n actual in that case then f of y n initial not equal to 0. So, this is the principle behind that suppose if you are guessing initially and if you are getting a function f of y n then that suggest that for the exact value of y n if for if your initial guess was an exact value of y n then f of y n would have directly yielded you 0 then there is no need of further solving the system equation. If your initial guess is not the actual normal depth y n what you can do is that you are now going to evaluate the function f of y n you are now going to evaluate this function f of y n subsequently you are going to evaluate f dash y n also you are going to evaluate f dash y n what is f dash y n this is d f by d y n. So, as taken from Hanif Choudhury 2008 on his book n flow through open channels. So, here d f by d y n you can give it as d by d y of a to the power of 5 by 3 p to the power of 2 by 3 minus j. So, this is nothing but 5 by 3 a to the power of 2 by 3 p to the power of 2 by 3 d y n minus 2 by 3 p to the power of minus 5 by 3 a 5 by 3 d p by d y n. So, you rearrange the terms recall that a by d y n in any cross section of the channel mini channel we had suggested that the top width d is nothing but equal to d a by d y. So, for the uniform flow d a by d y n will give you the top width of that uniform flow in that channel if you recall them we had discussed these things in the earlier classes. So, just substitute those quantity here if you substitute them appropriately you will get the corresponding relationship for the trapezoidal channel your f dash y n is nothing but equal to 1 by 3 r to the power of 2 by 3 into 5 times the top width minus 2 y's r d p by d y n and you also know that f of f of y n is equal to a r to the power of 2 by 3 minus j. So, what you can do is a f of y n is equal to a r to the power of 2 by 3 minus j. So, you have now two functions f of y n and its corresponding differentiated function f dash y n start with initial guess value for y n that is y n 0 is equal to some value some value you can specify you start with the initial guess value using your so using your y n value what you have to do is evaluate f of y n using your initial guess evaluate f dash y n using your initial guess. Now, your improved value for your normal depth y n so I am just giving it as super fix 1 the improved value this is nothing but the old value minus f of y n that has been obtained using your initial guess and f dash y n that has been obtained using your initial guess. So, this will give you an improved value for your normal depth y n so this is given as y n 1 once you obtain y n 1 you can check whether f of y n 1 whether it is equal to 0 you can check that if not then go for the next iteration that is in the next iteration y n 2 is equal to y n 1 minus f of y n. So, again check if your y n 2 satisfies that condition f of y n or using this y value if it satisfies that if it gets equated to 0 then fine that is the solution if not then go again for the next iteration y n 3 like this any general iteration can be given as i minus 1 minus f of y n using the i minus 1 this thing by f dash like this you go on. So, you can stop your iteration once y n converges what do you mean by converging convergence means you can put some convergence criteria such that between 2 iterations y n in the ith iteration minus y n in the i minus 1 iteration by y n in the i minus 1 iteration the mode of this thing if this difference is some less than some tolerance value tolerance value you can specify according to your thing some of them some people may specify 1 into 10 to the power minus 3 1 into 10 to the power minus 4 1 into 10 to the power minus 5 or whatever we according to the requirement you can specify any convergence criteria use them and see whether your y n is getting converged once it converges you can use that value of y as the normal depth fine. So, that is the method of using numerical methods to compute normal depth if you recall in the critical flow in critical flow computations we had we had used the concept called hydraulic exponent. So, the corresponding section factor for critical flow please note that this section factor is for critical flow this was given as z square is equal to some coefficient c into y to the power of m where y is your critical depth m was given as hydraulic exponent for critical flow. A similar analogous method is there for uniform flow computations also in uniform flow you have already seen section factor conveyance factor right. So, in uniform flow also the conveyance factor now can be given as k square is equal to some coefficient c into the normal depth raise to an exponent capital N. So, this capital N this is called the hydraulic exponent it is called the hydraulic exponent for uniform flow this you can use why this is being used it gives some certain characteristics of the channel you can it will be of aid if one know the hydraulic exponent of the particular channel sections and all it will be then quite useful to compute the normal depth means there is no need to further go and measure the depth or no need to go and measure the various other features using simple discharge and roughness coefficient you will be able to evaluate the normal depth. So, hydraulic exponent is also used therefore, hydraulic exponent is also used in computation of uniform flow how we arrive at that. So, in this thing I hope you everyone know that k is your conveyance factor n is your hydraulic exponent for uniform flow in this case I can just derive the following quantity k square is equal to c into y to the power of capital N taking logarithm both sides this is twice log k is equal to log c plus N log of normal depth if you differentiate this quantity with respect to your normal depth this is nothing, but N by twice y n just keep this as equation number 2 Manning's equation from the Manning's equation you remember the conveyance factor k is equal to 1 by N a r to the power of 2 by 3 use logarithm here differentiate that differentiating this thing you will get d by d y of log k is equal to this quantity differentiating this quantity in 0. So, you will see that this is nothing, but 1 by a a by d y n plus 2 y is 1 by 3 r d r by d y n. So, you know the quantity d r by d y n because r is equal to a by p this is you are quite aware. So, therefore, your d r by d y n term this is nothing, but equal to 1 by p d a by d y n minus a by p square d p by d y n again if you recall that d a by d y n is equal to your top with t use the following relationship go ahead your d by d y n of log k is now equal to t by a plus 2 by 3 p by a into t by p minus a by p square d p by d y n or this is equal to 1 by thrice area into 5 times the top with minus 2 r d p by d y n from this becomes your equations 3 comparing equations 2 and 3 you will get your hydraulic exponent for uniform flow n is equal to 2 y's y n by 3 a into 5 times t minus 2 y's r d p by d y n based on say based on cross sections whether it is trapezoidal whether it is rectangular triangular parabolic circular whatever be you can identify means you can find you can find n for each type of cross section fine you can find n for each type of cross section now for example a trapezoidal channel with bottom width b 0 side slope 1 is to be normal depth of flow y n what can you expect now in this thing again your area of cross section I can write it in the following form 1 plus b into y n by b naught r is equal to 1 plus b into y n by b naught whole thing again into y n by 1 plus 2 y n by b naught root of 1 plus b square top width t is equal to b naught into 1 plus 2 b y n by b naught isn't it these quantities you are already aware you can evaluate it on your own also using these things on substituting them in the equation for n hydraulic component exponent n I am not going to do it for you here you can do it as a homer you will see that your n becomes 10 by 3 into 1 plus 2 y is b y n by b naught by 1 plus b into y n by b naught minus 8 by 3 into root of 1 plus b square into y n by b naught 1 plus 2 y is root of 1 plus b square into y n by b naught like this you will get expression for n for trapezoidal channel. So, why I wrote n in such a form is that you can have a relationship of n versus non-dimensional depth y n by b naught you can plot them your plots may be you know how to plot by this time now. So, these figures if you see it may range from 2 to 5.5 and your n value may be something it is going like this for various cross sections. So, if you have any particular value say if you happen to observe n value for any particular channel say it is here then you can just check the corresponding depth of flow you can interpret that and from that you can get your normal depth of flow. Similarly, if it is here corresponding thing you can easily interpret them. So, it is up to you to determine them. So, that is you can use the normal you can use your hydraulic exponent for uniform flow you can compare it with the non-dimensional normal depth of flow use those graphs and interpret the normal depth of flow. So, next we are going to deal with is what happens to channel flow or uniform flow if composite properties are there if composite properties are there what happens to your channel flow. First one if your channel section if your channel section is having composite roughness if it is having composite roughness what happens to your uniform flow see it can be say for example, in the trapezoidal channel the side walls may be of one particular material the bottom bed may be of some material this may be of another material still the uniform flow is there in the channel how will you compute uniform flow it is quite difficult or it is tedious if you then take the corresponding areas allotted to these perimeters and try to evaluate it independently it will be quite different finely to get your normal flow normal or normal depth. So, how will you evaluate the uniform flow in such situations this is different this is different this is different. So, you can just suggest now that the wetted perimeter is of different materials say or for example, another simple case is in a rectangular channel the side walls are made of glass and the bottom is made of wood what happens to your uniform flow how will you compute them. So, in these situations you require a concept called equivalent roughness coefficient. So, we can use the same Manning's equation to compute the uniform flow, but using your equivalent roughness coefficient. So, how will you evaluate equivalent roughness coefficient what is the procedure you may or may not divide it the areas wherever the properties are changing corresponding perimeters say cross sectional area divided into sub areas into n sub areas each having wetted perimeter p i its roughness coefficient n i. Now, we are assuming we are assuming that here the channel cross section it is not that much the roughness things are the roughness is the thing that is getting default here. So, we are assuming that the velocity the average velocity in each section each sub area is same as that of average velocity for entire flow if we assume that. So, that means, that say v 1 bar is equal to v 2 bar is equal to v 3 bar equal to v i bar equal to v n bar and this is this are all equal to the entire average velocity of the cross section of the channel then you can evaluate the equivalent roughness coefficient in the following form. This is one particular formula i is equal to 1 to n p i n i to the power of 3 by 2 by i equal to 1 to n p i whole to the power of 2 by 2. So, students please note that the capital N used here is not your hydraulic exponent it is just to show the summation the total number of sub areas divided in the channel cross section it is just to denote that you can also evaluate equivalent roughness coefficient by assuming that means you can assume that the total force resisting the flow in the channel cross section it is equal to the summation of the forces individual forces that are opposing the flow in individual sub areas that can also be used. So, if you use that method you can give your equivalent roughness coefficient n e is equal to i is equal to 1 to n p i n i square p a whole to the power of half this is one method. Suppose in some other literature you may see some scientist they might have used that the total discharge in the channel is equal to the summation of the discharges in individual areas if that concept is used then your equivalent roughness coefficient n e can be computed as p r to the power of 5 by 3 where p is the total wetted perimeter of the cross section r is the hydraulic radius for the entire cross section of the channel then i is equal to 1 to n p i r i to the power of 5 by 3 by n i like this you can evaluate it. So, you can use also this particular formula once you get your equivalent roughness coefficient you can use it to compute your average velocity r to the power of 2 by 3 s naught to the power of half you can also use it to compute your discharge a r to the power of 2 by 3 s naught to the power of half. So, we will conclude here. So, we have discussed on the how to compute the composite or effective roughness coefficient or you can tell it as the equivalent coefficient of roughness to compute the roughness coefficient if there are different surfaces or if there are different surfaces in the channel cross section you can use them to evaluate the Manning's velocity and discharge next time means we will conclude the portion here next time we will be dealing with how to evaluate uniform flow if there is difference in cross sectional areas if the shape of the area and all if it is quite different how to evaluate the uniform flow. Thank you.