 we need to see that how we can simulate these wave flow and then how we can take protective measure preventive measure. So, for all those understanding wave is very important and as such we started with what are the different types of wave how we can classify waves and then as I am just saying that the how the wave propagate that is very important for our various requirement practical requirement that is why the speed of wave propagation is also necessary and there comes the term that we call as celerity well what we mean by celerity is that it is the speed of wave with respect to the fluid media well if we just simply say that celerity is the speed of the wave then it is not complete because we need to say that it is in reference to the fluid media if the fluid media is it at rest then of course the speed of the wave absolute speed of the wave and celerity will be same but if the fluid is also moving and then over the fluid the wave is moving then celerity is the speed of the wave with respect to the fluid media and for analyzing these wave say what is the celerity to have an expression for the celerity when we try to analyze it at unsteady flow we face some complexity and this unsteady flow therefore we try to convert to an equivalent steady flow so that it become convenient for our analysis and last class we did so how we can we could see how we can convert this to a equivalent unsteady flow and we are starting from equivalent steady flow and we are starting from that in this class well say this is a wave forming here a solitary wave single wave and the bed is there and by say and let us first consider that this is a steel water maybe in a tank and by moving a paddle we have created a wave a single wave is moving like that and it is moving with a speed c well and this is a unsteady flow this is a unsteady flow now to convert this to a steady flow situation as I did explain in the last class that like a walker as the belt is moving upstream or say when we walk the belt moving the opposite direction in a walker and that is why although our that we are walking on a walker but we are not moving forward and then someone can observe us that how we are behaving when we are walking but we are in a steady position we are in a static position so that advantage we can have here like the belt moving the opposite direction in a walker we can put a opposite direction the velocity c which is the actual velocity of the wave then what will happen that this is also moving in a velocity c wave is also moving in a velocity c in this downstream direction and as now this fluid media is moving in the opposite direction with velocity c and wave which was moving in the downstream direction with a velocity c so relative velocity of this will become 0 and then it will be remaining in this position and now we can observe it carefully well now let us see how we can analyze it say this depth p at this section suppose at this section depth b y and so amplitude of the wave is say a this is the amplitude of the wave and so what will be the depth here depth in the wave portion where we have the wave this will be say y plus a well and when we talk about velocity then c is the velocity at this section say section 1 and if I talk about the section 2 then what will be the velocity say velocity is v because this velocity will not be equal to the velocity c because depth here is change so let us write this velocity as v well now we can write a continuity equation and from that continuity equation we can find that how we can express this v and c how we can relate this v and c well how we can write the continuity equation say if we consider this channel to be say rectangular and suppose we are considering unit width of the channel so this is let me just write here this is a equivalent steady state and now let us see the continuity equation how we can write say continuity equation between section say 1 and 2 between section 1 and 2 how we can write and that we are writing considering unit width of the section unit width of the section well we are considering unit width of the section then what will be the discharge in continuity equation state that here the discharge and here discharge will be equal discharge at section 1 and section 2 will be equal fine now discharge is equal to the velocity at this point 1 section 1 and the area area is equal to y into each width in fact if we consider width then width will be coming but we are considering width to be equal to 1 so it will be c into y so we what we can write that c into y is equal to here we are talking the velocity as v so v and depth here is equal to y plus a and from this what we can write that v is equal to c y divided by y plus a let us keep this expression here and now we can write another expression this is from continuity equation we are getting this relation that is v we can express in this form well but we can have another expression from the consideration of say energy equation now what will be the energy at this point it will be y 1 plus that means at section 1 it will be y 1 plus sorry y we are writing y y plus c square by twice z and here because velocity is c here and here it will be y plus a plus v square by twice c and let us write that expression well and you can write that energy equation energy equation between section 1 and 2 of course when we are writing energy equation between section 1 and 2 and we are saying that energy at this section at that section are equal what we mean that there is no energy loss in between that assumption is inherent there well now let us write that expression say y plus c square by twice g that means v square by twice v and velocity square by twice g this is equal to depth in the section 2 is y plus a plus say v square by twice c let me write one step v square by twice c now what is v is we know that v is equal to c then y by y plus a so we can just replace it in this form that is y plus c square by twice g is equal to y plus a plus write that c square this in place of c square by y square by y plus a whole square so v square we can write that v square by twice g will be and v we can write c square into y square by y plus a whole square well this is the expression and from this let us try to write because our interest is to get an expression for c and here what is not known and what we want to try is the c and what is our unknown thing is y a here the amplitude and at the same time this expression we are writing for a very for a way where the amplitude is very small I mean small amplitude way as we just if we recall our last class then we were talking about one expression that c is equal to root over g y for rectangular channel we obtained that and that was for small amplitude way so here also we are considering that this depth is considerably large than the amplitude a well let us just separate this term and see how we can write it this y and y will of course get cancelled and let me write first a in terms a is equal to then c square by twice g c square by twice c is here so we can bring c square by twice g and then we can write one minus say y by y plus a whole square well and this can be written as say let us break up y c square by twice g our intension is to see that if because in the expression of c in fact we do not have the a term root over say g y that was that that what that is what we got in the last class so let us see by we are starting from the first principle and we are trying that yes if we can get this expression and here we do not have a a term so let us see how we can do that and this we can write say y plus a whole square and then this will be say we can break up y square plus twice a plus twice a y y square plus twice a y plus a square then minus y square well here we are in a position to simplify now well so what we can do that from here let us write one expression for c so c is equal to we can write that c is equal to twice g a will be coming c square let me write c square is equal to twice g a and then this will be just we can write in the reverse direction this will be say y square plus twice a y plus a square this is there and here what we will be having this and this is getting cancelled already and here we have twice a y plus a square so that twice a y plus a square we can write here as a keeping common and we are writing twice y plus a well now this a and this a we can cancel at this point and let us see how we can write this expression well so we can write c square is equal to now root over twice g twice g and entire things we can put under root sign root over twice g and then here it will become say y square plus twice a then twice y plus a let me write c square is equal to now root over twice start from this expression again y square plus twice a y plus a square divided by a twice a plus it is a and a is getting cancelled so what we are writing that twice g and in this part we are writing that expression well plus a square well so now here as we know that a square is say a itself is very small so this a square term we can put equivalent to 0 or say it is negligible so if it is negligible now what we can write that c square now it is c because we have written already in terms of root over so c is equal to say our target is to get this term g y so we are writing g y means from here we are bringing a y common we are bringing a y common here in fact we had a term a y a y plus a square right so y we are bringing common and 2 we are taking to that side so what we can write root over g y then here we can write so as 1 y has come so it will become twice y 2 is going inside and then plus 4 a 2 is going and this is coming so it will be twice y plus a well now this we can further simplify in a way that this is equal to root over g y and let me write it in this form that twice y plus a fine here also let me write twice y plus a then in fact we need to write plus twice a we need to write plus twice a well so this twice y plus a twice y plus a that we can have 1 and this will become twice a by this expression and so let us write c is equal to root over g y then we can write root over 1 plus twice a by twice y plus a now this expression twice a so a is very small and here in the denominator we have twice y so it is quite a large expression and as y is very large as compared to a so this expression will become every small term divided by every large term so this we can now neglect so if we neglect this term say as a is very very small as a is very very small as compared to y therefore 3 a by twice y plus a is say negligible is negligible and now just neglecting this part it will become c is equal to root over g y well so that is the expression that we can derive what we got in our last class again starting from the first principle we could see that for a small amplitude wave we can have the expression for c as root over g y and if it is a suppose non rectangular channel then what we do that y will not represent this things properly so what we write that c is equal to root over g d what d is d is the hydraulic depth d is the hydraulic depth so using this hydraulic depth in fact for rectangular channel this will become y but for trapezoidal channel it will be a by t so using this expression also we can get the value of c so after that let us see that well for a wave of this type for a wave of this solitary wave of this solitary wave of this kind we can have an expression for c but if we have a charge if we have a charge and how to analyze this charge whether the c value will be same as or the same expression whether we can use or not well that first we should know what is charge this is also in unsteady flow this is important I mean important topic that is charge we need to do lot of analysis in fact on charges and but here we will be just doing some limited exercise on charge that you can see that charge is a moving wave front that brings about an abrupt change in the flow depth of the channel so basically what we mean by charge is that say this is the bed and normal flow of charge is suppose moving here and somehow a wave is coming like that it is a moving wave front that is why we call this as a wave front this wave front and then it is so this water is moving like that otherwise also and then we have a moving wave from this wave front is moving with a velocity say v w this wave front is moving with a velocity v w and this can be this we refer as a charge in fact it can be in the other way also say your water level is here water level is here and water itself is moving in this direction suppose and somehow due to closure of gate the you have suddenly closed the gate so water level here is rising and then the wave can be moving in the opposite direction also this is say wave moving upstream and this is a wave moving downstream so whatever may be the situation we call such moving wave front which is moving and creating a abrupt change in the depth so depth earlier was like this and there is a abrupt change if the depth changes gradually like this then we will not call this as a charge but here depth is changing abruptly and that is why we call this as a charge and the charge and so this charge also we can express in the form of say equivalent steady state for its analysis well let us see how we can represent this in the form of a equivalent flow well let us consider this one anyone we can consider but let us consider this one suppose this is say one and this is section two so say one and two why we are considering because this is both in the same direction here it is in the opposite direction that will give us an opportunity to handle a case where the flow and waves are moving in the opposite direction till now we have not discussed that sort of situation so suppose the velocity here is v1 and velocity here will be changing velocity here means in the entire section say v2 well now vw is the velocity which is moving in the opposite direction this is a unsteady situation at the next moment of time this wave will be moving further this wave will be moving further and say we will not find it here so let us see how we can do this to keep in a steady position well what we can do in this we can apply an additional velocity in the opposite direction to the speed of the wave well so please just see we are meaning that opposite direction to the speed of the wave but this speed of the wave is not the celerity celerity will be the relative speed so we are we need to apply a velocity in the opposite direction to the speed of the wave otherwise it is moving with a velocity v1 now if we apply another velocity say vw if we apply another velocity say vw in this part then total velocity at this point will become vw plus v1 and here also it will become v2 plus vw and then velocity of this in this direction will become 0 because it is moving in this direction it is and then that is why the this things is not in a position to move further it will be remaining in the same position well so let us represent this and let us see how we can derive okay I am now having this portion and this equivalent steady state position we can keep here we can draw here so it will be like this now we have here the velocity will be v1 plus vw and here the velocity will be v2 plus vw and of course there will not be any velocity of the wave it is not moving rather so this is basically we can say equivalent steady situation well now again for this search just like we were writing earlier we can write continuity equation and momentum equation so continuity equation we can write here say continuity equation and let us put some depth value here say depth here is y1 and depth here is say y2 well now what is continuity equation so again considering unit width of the channel considering unit width we should mention this one considering unit width this will become say y1 into v1 plus vw plus vw is equal to y2 into v2 plus sorry it is not v3 it is vw v2 plus vw well so that is our continuity equation and this continuity equation can give us or from this continuity equation we can express y1 in terms of y2 and we can do those things and then what is this v1 plus vw that is also interesting in fact if from this point when wave speed or absolute speed of the wave is say it is moving with a speed vw and then the fluid is moving in the opposite direction v1 then what is its relative speed its relative speed will be v1 plus vw v1 plus vw because both are moving in the opposite direction so it will be it is in fact it will appear that it is moving faster when v1 is in this direction vw in this opposite direction so wave speed relative wave speed will be v1 plus vw and this relative wave speed in fact we call as c so this v1 plus vw is nothing but c well that we can write that equation so continuity equation we have seen and then let me write the continuity equation again and let me keep the diagram here quickly say it is v1 plus vw sorry v2 plus vw and this is v1 plus vw that we have and depth here is y1 this is y2 now the continuity equation we have already described so y1 plus y1 into v1 plus vw is equal to y2 into v2 plus vw now we need to write the momentum equation we need to write the momentum equation so for momentum equation for writing the momentum equation we need to see what are the forces acting so the pressure force from this direction and the pressure force from that direction and then the momentum change in momentum between this point and that point so pressure diagram we can draw like that and as we know we have discussed that part a lot so pressure here we can write as half of rho g into y1 square because pressure here is nothing but rho g into y1 or w y1 we can write unit weight as w or we can write rho g so writing in any form we can get it say rho g y1 and then area of this triangle will be half of rho g y1 plus multiplied by the way y so it will become half of rho g y1 square so momentum equation what we can write the force from this side that we can write this is continuity equation and now we are writing momentum equation so momentum equation we can write here say half of rho g y1 square that is the pressure force from this side and minus half of rho g y2 square that is the pressure force from that side that is equal to change in momentum rather rate of change in momentum so what we can write that mass flowing per unit time and change in velocity so mass flowing per unit time how we can write that discharge flowing per unit time well discharge as it is not changing from here to here no other flow is being added nor the flow has been taken out so we can write it in terms of discharge of this force because this is known to us what is depth y2 that is of course not known to us because after coming of the wave this is happening but this y1 will be generally known to us so let us write the discharge in terms of this one that we can write discharge in fact is equal to y1 into v1 plus vw is the velocity and area is actually y1 into width but width is 1 so this is what the volume flowing we can say and then this volume flowing if we multiplied by rho then it become the mass flowing so this is what the mass flowing and then v square or say sorry not v square this is the mass flowing and the change in velocity so change in velocity is equal to in fact v2 plus vw minus v1 plus vw so ultimately the resulting will be v2 minus v1 so that is what the change in flow velocity this is our momentum equation well from this momentum equation we can further derive some of the relation which is required for finding the expression for c in case of charge and well before that let us write the expression for c here this is another equation that is c is equal to we have we can write that c is equal to say v1 plus vw that we have already explained how it is so this is equal to c celerity c that is the expression for celerity c and this expression celerity c we can further say starting from the momentum equation we can derive it in a different form and let us see how we can do that well now we need to mention one point here that we got one expression for c is equal to root of our gy and this equation is called Lagrange's equation of celerity this equation c is equal to root of our gy this equation is called Lagrange's equation of celerity well and now let us see what will be the expression for c in case of charge in terms of y1 and y2 well in general we can have it as v1 plus vw but we then we do not know what is vw and so we need to have it in a different form like that earlier we got in terms of depth y so let us see what we can do starting from this again starting from this momentum equation we can write that this is equal to say rho we can cancel in each part then let me take common here say half of g then y1 square say half of g y1 square if I keep here then it become 1 minus y1 square by y2 square well and on this side we can take say y1 is here and then v2 minus v1 is there so what is the v2 is the velocity so velocity and this expression actually y1 plus v1 plus vw this expression this expression is nothing but this is showing how much is the discharge I am writing this as q this is q means unit discharge because we are talking about discharge through unit width so this is equal to q and what is our v2 or v1 v2 is nothing but q by say y1 because v2 into y1 is equal to y2 is equal to q and v1 is equal to q by y1 because that we know that is again from continuity equation we know that v1 y1 is equal to v2 y1 that is also known to us so in fact this v2 can be written as v2 can be written as q by y2 means q is equal to v1 plus vw q is equal to y1 v1 plus vw divided by y2 so that can be written so well so if we bring this q common here let me write it in this from here say y1 plus v1 plus vw and then this q we can let us keep rho still here because here also rho discharge will be coming so this discharge is nothing but say we can again write this part as y1 into v1 plus vw that is what the discharge we are writing here bringing common and then inside this will be say 1 by that means q we are bringing out this q is nothing but this expression and then here it will be 1 by y2 minus 1 by y1 and as we can see very clearly that this term and this term is same so we can just because we are writing for discharge only so this will be square so this can be written as half of gy1 square and 1 minus y1 square by y2 square is equal to this now we can write y1 square and then we can write here as say v1 plus vw square then within bracket we are putting say 1 by y2 minus 1 by y now let us see how we can further simplify this expression here that is the v1 plus vw is nothing but c so from that we can put the value here as c square say y1 square and this y1 square we can cancel and then we can just rewrite this expression in a simple form like that c square we have already replaced this part that means this we are writing as c this one this is nothing but c so putting that what we can write c square is equal to say half of g half of g then we can write say y square y1 square minus y2 square that we can write y1 plus y2 into y1 minus y2 and this is divided by y1 square and then this the denominator part we can write say y1 minus y2 divided by y1 y2 that means this part this is become y1 y2 then y1 minus y2 so this is coming here and this is becoming y2 square and y2 square then this y1 y2 square minus y1 square that we are writing as this one say y1 y2 and y1 plus y2 now from here what we can do this is equal to half of g then we can write y2 by y1 so this y1 and this y1 is getting cancel and we can write this as y2 by y1 y2 by y1 y2 by y1 into y1 minus y2 by y1 minus y2 by y1 minus y2 by y1 say which will be y1 plus y2 that will be the expression and we can of course further simplify from this part you can see that our target is always to get the expression in terms of root over gy and then whether we can have it in further simplified form so c square this is equal to we can bring one y from here one from one y one from here so this is equal to say gy1 one y1 we are bringing from here and then the other part we are writing in a different way say half we are writing here and this y2 by y1 is there y2 by y1 and then as one y1 we have brought here this will be 1 plus y2 by y1 well so that is the expression we are getting and as you can see that this will be y1 so c square is equal to this part then if we just try to take c then we will be getting c is equal to root over gy1 then root over say half of y2 by y1 and 1 plus y2 by y1 plus y2 by y1 well now in this expression normally if we refer to our earlier diagram this say y2 will not be known to us because it is coming this wave is coming and the y1 will be known to us of course so to get the value of c then say generally we are getting the expression in terms of y2 by y1 and here it is gy1 and this y2 we need to know in many case so this is y2 sorry this should have been y2 square by y1 square yes we are taking common y1 square here and as such it will be y2 square by y1 square and this is y2 square by y1 square right and that is why we are getting this expression in this form like that one this one and so this is coming in this form well fine so that way c is equal to root over gy1 and this expression we are getting in terms of y2 by y1 and this is the expression normally we use for computing charge ok charge means a celerity in charge and then this help us in various in solving various numerical problems small numerical problems we can solve by using this particular expression well at minute for c2 however continuity and momentum equation and that is what we need to know so we are getting this expression in terms of y2 by y1 39 is not it 39 means in other words it takes 20 minutes 19 minutes problem is not going to happen in 19 minutes not at all it is not that easy right? it is not easy it is not that easy it is not that easy let us take a problem typical problem in fact the problem, typical problem, in fact we will not be solving the problem as such, but we will see how the steps need to be followed and how we can get the results, so, the problem suppose he is like that, what are flows in a rectangular channel at a depth of 2 meter we carefully 1 meter per second, well, say this is the channel and here water is ంిితి బానిమి మిడ్చన్చి వటం్సిస్ని పరపటిత్బిమాసి మారందాహి మరనిమలికిసంచె. So, this is say depth is 2 meter and the velocity of flow is 1 meter per second well and if discharge in the channel is suddenly trivial what will be the depth of flow. So, here it was the depth of flow suppose in the discharge if we suddenly trivial then what will be the depth of flow suppose depth of flow is increasing like that and this will in fact continue to increase this will continue to increase this will continue to increase and then wave is moving in this direction and after sometime the depth will be increasing. Now here this is in fact our say we can put as y 2 and this is our y 1 of course we can write either way also and then velocity here is say v 1 velocity here is v 2 and then wave velocity and then wave velocity let us put as v w. Then how we can write continuity equation and momentum equation for this case and what we know here is the depth y 2 and this v 2 is known velocity is 1 meter per second and depth is say 2 meter. So, that part is known and other things are not known and what we are knowing the discharge what was earlier flowing is just trivial. So, this discharge what will be getting here y 1 v 1 say considering unit width what will be getting y 1 v 1 that will be 3 time of that of the say y 2 v 2. So, how we can write the continuity equation again to first make it steady what we can do to make it steady we can apply a velocity opposite to this v w. So, we will be applying here v w and then if we apply v w what will be the velocity here ultimately this velocity will be changing to say v 1 minus v w after applying this one and this will be changing to this velocity will be changing to v 2 minus v w well and then continuity equation we can write say v 1 minus v w into y 1 is equal to v 2 minus v w. So, that is what we can write the continuity equation now here we can put the value of v 2 and y 2. So, what we can write that this is nothing, but say v 2 is equal to 2. So, 2 sorry v 2 is equal to 1 1 minus v w into 2. So, this equation will give us some information then we can write the momentum equation we can write the momentum equation say. So, momentum equation can be written as say half of well let me give a number to this equation say this equation is 1 then momentum equation is equal to half of rho g say y 1 square minus y 2 square is equal to rho v 2 minus v w here this is the velocity then we are getting y 2 and I mean this part is actually the discharge and then it is v 2 minus v 1 that is what we are getting. And then this expression also we can simplify means we can put the value like half of say rho g y 1 square minus y 2 square means 4 this is equal to rho we can write this as rho then v 2 minus v w that we can again write as say v 2 was our 1 minus v w and this y 2 is already known to us. So, this will this is 2. So, this 2 depth is we got the depth s sorry we got the depth s 2 meter. So, that we can put here say 2 and then this v 2 minus v 1 again this we can write as 1 minus v 1. Now, this is say equation 2 well that what information we have we have a relationship between what information we have that you can refer to the slide we have one information like that v 1 y 1 which is the discharge which was the discharge after the charge is coming that is v 1 y 1 this is the actual discharge what is coming here and this discharge is triple or three times of that v 2 y 2. So, we can write it like that v 1 y 1 is equal to say three times of v 2 y 2 and this value we can have that is equal to three into say one is the velocity into the depth 2. So, this is equal to six. So, that way we can have I am just writing the numerical value not writing the unit here. So, this is one expression and then we have these expression and solving these expressions we can get ultimately the value of y 1 because our target is to know the value of y 1 and. So, if we know this value of y 1 our problem is solved and to get the value of v 1 we have these equations say this is one equation and then here of course, we have this unknown v w and this is one equation where we have this unknown v w. So, v w and y 1 are the unknown that we can solve by trial and error method and we can get say y 1 is equal to 2.45 meter. This is just a solved problem already in the book of Rangaraju and just I am taking here just to show how the search problem can be solved. Well, then we will be taking up the next topic rather the equation of unsteady flow. Well, well, so equation of unsteady flow that we know that already we have discussed some of the equations of unsteady flow when we were discussing our earlier discussion on say governing equation of a functional flow. For unsteady state we were discussing about the continuity equation. The continuity equation that as we know it is say del a del t plus del q del x is equal to 0 that we could get and this of course, we can write as say del a del t plus this q if it is q is written as a into v this can be written as say a del v del x plus v del a del x is equal to 0. Well, so here I am not taking up again this continuity equation that you can refer or rather we will be knowing by referring to our earlier classes and then the two governing equations are required basically for solving the problem of unsteady flow. And then one of the most popular equation of course continuity equation is there then another equation that we take is the equation of motion and this can be of course, we use momentum principle or energy principle to get this equation of motion and for some case we use say continuity and momentum equation in couple for solving the problem. Sometimes we use continuity and energy equation for solving the problem depending on the situation where suppose energy is energy loss is there there we go for momentum couple continuity momentum couple. So, that way these considerations are there of course, in this class we are not going for all those detail, but we will be just discussing the equation of motion that was given that was first deduced by Barrett Saint-Venant and popularly known this equation as Saint-Venant equation and that is say 1879 long back this equation was derived well and this is a dynamic equation of unsteady flow in open channel and for what condition it is the shallow water wave equation shallow water wave equation. That means why we are giving this a shallow water wave equation because when this equation was derived this vertical acceleration was considered to be negligible. So, vertical acceleration is negligible and then we should know that for what condition we can apply this and derivation of this equation not this equation this is continuity equation derivation of the equation of motion can be done by two approach one is energy approach and another is that we can start from Newton's equation of motion. So, by both these approach we can have the derivation and here we will be discussing how by energy approach we can derive the equation in a simple way and then let us see what is the energy equation say if this is the flow and this is going like that well let me take some space because we will have to write here. Say this is section one and section two so section one and two well in this section one and two let us take a small length let us take a small length and let us write this as say as this small let us write this as say del x this length is and datum is here datum is here. So, this depth is say z one and this depth is say z two that is the elevation head up to the bed and then up to this depth this is y one we are considering slope to be small that is why we are not writing y one cos theta or those term we are writing straight weight is y one and then here this energy is v square by twice g v one square by twice g energy head up to that point and then here in the second section section two this is the y two and the v square by twice g is suppose that much v square by twice g v two square by twice g and then we can draw the energy gradient line as this one energy gradient line as this one and there is definitely some energy loss this we refer as so when we draw it parallel to the datum then we will be finding that there is a head loss and this head loss we can write as h l and in fact this head loss is having two component one is that one is that it is friction loss that is known to us because we did discuss about this friction loss during our discussion of gradually varied flow itself and but here there will be another component because in case of unsteady flow that local acceleration exist local acceleration exist just to recall briefly say when we say v then it is a velocity which is a function of x and t of course in steady flow it become a function of it is not a function of t but in unsteady flow this become a function of t now when we talk about so this change in v say d v that we can write as del v del t into d t plus del v del t into del v del x into d x del v del x into d x now if we write that it is a function of x and t so if we just write it as d v d t d v d t then we are getting del v del t plus say this d x d t so d x d t divide into del v del x and this d x d t is nothing but v so now perhaps you can recall that we did discuss earlier also d x d t is equal to v del v del x and then this is called local acceleration this is called convective acceleration now in case of say steady flow this term is not existing so that problem become little simpler but here we have this term so this local acceleration exist so because acceleration is there in any whether it is fluid or any solid but if acceleration is there that is a mass moving and that mass is having some acceleration means some force is being utilized to have that acceleration otherwise that acceleration will not be there so some force is being utilized to have this acceleration and when some force is being utilized and things are moving then there must be some work done force is applied and it is moving so some distance has been moved so some work done will be there and some work done is there means some energy loss is there so we can say that in this sort of flow there is two type of loss one is h f another is h a so this two loss will be there this h a is say loss due to acceleration well now if we write the energy equation so writing energy equation between say 1 and 2 between 1 and 2 what we can have between 1 and 2 this we can write say z 1 plus y 1 plus v 1 square by twice g is equal to z 2 plus y 2 plus v 2 square by twice g plus h l well plus h l or say we can say loss loss h a loss loss h l okay now this loss as I have explained that let me write here say loss h l is equal to say friction loss friction loss that we write as h f and plus the acceleration loss acceleration loss that is say h a so our expression become now of course in this loss term we need to put this two so this is equivalent to h f plus h l already we have written here now we need to have an expression for this two so h l is equal to we know that it is h f plus h a now how we can get the h f because friction slope we know that as f if friction slope is as f friction slope is say as f then we know that as f that is friction loss divided by delta x delta x is equal to as f or we can say that as f the slope into distance as f into this delta x will give us the h f so friction slope may be this like that and we can have as f up to certain point h f and then this slope is actually as f and then other part of loss is due to say this acceleration so that is clear to us that we have discussed earlier also so in place of as f by delta x also in place of as f we can write that as f is equal to as f into delta x well now that so for our next acceleration loss is there how we can calculate that so acceleration del v del t local acceleration del v del t exist here so if this is the acceleration then if this is the acceleration then the mass suppose if we consider that unit weight of fluid unit weight of fluid then mass is equal to say weight by actually weight by the g so this is here as our weight is equal to 1 so this is equal to 1 by g and force utilized forced utilized to produce this acceleration force utilized to produce to produce this acceleration that is equal to mass into acceleration mass into acceleration so that we can write as 1 by g into del v del t so this is what the force required force utilized then this forces work for how much distance this force is working and then acceleration is being there and then the fluid is moving this mass is moving from here to here so the distance of say delta x it is being moving so what is the work done and distance move distance moved by this unit mass by this unit weight rather by this unit weight of fluid is equal to delta x that is what we are considering therefore the work done how much work is being done work done is equal to 1 by g del v del t into del x del v del t into del x well so this is the work done and what this work done is that is the energy loss energy loss so energy loss is equal to that means 1 by g del v del t into delta x so finally how we can write the head loss so this energy loss is nothing but h a so this energy loss due to acceleration that is the work done due to produce that acceleration and that is the energy loss we can write like that therefore so what we can write that head loss h l is equal to h f what we can write h f plus h a h f plus h a and that h f we can write s f into delta x and then this h a we can write that is the acceleration loss 1 by g and del v del t into delta x so putting this value of h l in our equation of energy this one then we can get a simplification of this equation of this equation of this equation of motion and that we will be doing in our next class and then from that we will be also trying to see how this equation can be solved and then by solving this equation how we can get solution to our real life problem of open channel flow so thank you very much.