 Myself, Deshmukh Sachin working as an assistant professor in civil engineering department of Oval Chain Stock Technology. Today, we are going to learn about a rapidly varied flow. After completion of this topic, you are able to understand what is the concept of rapidly varied flow. You can derive the equation for conjugate depths or we can say sequent depths and also you can solve the problems on conjugate or sequent depths. First of all, we will just see what is a rapidly varied flow. Last time we have seen the gradually varied flow where the parameters are changing very gradually. Now here, it is changing abruptly or you can say rapidly. The rapidly varied flow phenomenon is a very unstable and it has an inherent tendency to change itself to the stable tranquil flow resulting in the abrupt rise in the elevation of liquid surface. The local phenomenon where the elevation of water surface is suddenly rises is called as a hydraulic jump or a standing flow. Let us see this figure. The flow is coming like this. Here it is a barrier and it is coming abruptly down and it flashes again coming up. See here, this is the water surface which is coming down, this is the collision and it rises up. The rapidly varied flow we can observe over here. The rapidly varied flow we are observing over here. Here gradually, it is changing very slowly, very slowly, very slowly. Here we can say it is a uniform flow and this is a non-uniform flow up to where it is going to become a stable one. Now as the example is hydraulic jump of the rapidly varied flow, we will see what is a hydraulic jump. When a super critical flow where the Froude's number is greater than 1 changes suddenly to the subcritical flow where the Froude's number is less than a hydraulic jump, less than 1 in the hydraulic jump. Why Froude's number we are taking? Because Froude's number is the ratio of inertia force to the gravity force. Due to the formation of hydraulic jump, large scale eddies are formed and the turbulence occurs resulting in appreciable loss of kinetic energy takes place. Bidon, it is the name of the scientist, a French engineer, was the first to investigate the phenomenon of hydraulic jump experimentally in the year 1880, 1818. This is the same figure where you can observe the sudden rise over here, sudden rise you can observe here in the section 1 and section 2 and particularly here we are going to adopt instead of energy equation where taking a momentum equation, keep in your mind. Now define a hydraulic jump, the answer is when a super critical flow changes to subcritical flow, a considerable amount of energy loss takes place with this phenomenon is termed as a hydraulic jump. Now as appreciable loss of energy changes occurs in a very short length, this is the way of the channel, it is advisable to apply a momentum equation instead of energy equation along the continuity equation for the analysis of hydraulic jump. As per Newton's second law of motion, algebraic sum of all the external forces in any direction is equal to change in momentum flux in the same direction. Therefore applying the equation in the direction of flow to the control volume, this section, equation between 1 and 2 is called as a control volume, we will get summation of effects that is summation of all the forces is equal to mass of the flow into change in the velocity. So it is rho mass density into discharge into velocity that is final minus initial velocity. From figure rho q v 1 v 2 are the density discharge and initial velocity and final velocity respectively. Now the summation of effects is f 1 minus f 2 plus w sin theta minus f 4 minus f 5. See here f 1 it is in the direction of flow, f 2 opposition f 4 and f 5, this is a bottom resistance, this is the air resistance and w sin theta is with respect to this flow. Where f 1 and f 2 are the pressure forces at section 1 and 2 that is from the figure w sin theta is a component of the body force in the direction of the motion and f 4 and f 5 are the channel bottom resistance and air resistance. Generally f 4 and f 5 are very small, we are neglected and the channel bottom being practically horizontal the term w sin theta is also neglected. So solving these we will get f 1 minus f 2 is equal to rho q 2 v 2 minus rho q 1 v 1. We can solve it step by step, you will get f 1 plus m 1 is equal to f 2 plus m 2, where m 1 that is a mass at section 1 that is rho q 1 v 1, m 2 is rho q 2 v 2. Dividing the equation throughout by gamma that is we can say in many of the cases we are taking as w or we can say gamma that is specific weight. If you can divide this you will get again simply by it slowly dividing you will get this equation f 1 plus m 1 by gamma 1 is equal to f 1 plus m 2 upon gamma. That is specific force before hydraulic jump is equal to specific force after the hydraulic jump is known as momentum equation for an open channel flow. This particular equation is very basic equation or important equation to derive the equation for the hydraulic jump. F s that is a specific force depends on the depth of the flow cross section of the channel and discharge passing through the channel. The other equation can also be written as a 1 y 1 bar that is y 1 bar plus q square upon g a 1 is equal to a 2 y 2 bar plus q square upon g a 2. This is a very important equation. This equation is called as a momentum equation for an open channel flow. Now we will see what is the hydraulic jump in a rectangular channel? What is the equation for this? For a frictionless rectangular horizontal channel the momentum equation is the same a 1 y 1 bar plus q square upon g a 1 is equal to a 2 y 2 bar plus q square upon g a 2. This we can write as b is the width and y is the depth. Simplify it b into y 1 that is the area of 1 and remaining as it is a 2 is b into y 2. Put these values over here wherever area is there you can go for width into depth where y 1 and y 2 are the depths of the flow at section 1 1 and section 2 2 respectively. That is before and after the hydraulic jump and b is the width which I said to you therefore dividing the whole equation by b you will get this equation. Also we know q by b is a small q that is discharge per meter width. Putting these values q square for q by b that is capital Q by b square in the above equation put these values and solve this quadratic equation. Solve this quadratic equation once in y 1 and once in y 2 you will get y 1 square y 2 plus y 1 y 2 square minus 2 q square upon g is equal to 0 or y 1 y 2 square plus y 1 square y 2 minus 2 q square upon g is equal to 0 which is in y 2. Then simplify it for y 1 you will get the equation y 1 upon y 2 is equal to 1 half into bracket minus 1 plus under root 8 f r 2 square. Similarly the following equation we can write as with respect to the Froude's number where y 1 and y 2 are known as the conjugate depths or the sequent depths known as the ratio of these two. This is also relation between the pre-jump and the post-jump. Here the important thing that is q square upon g y 2 cube that we have to insert with respect to the derivation while deriving this while deriving this we should we should also focus on this what the quadratic equation and find out the ratio of these conjugate depths y 1 upon y 2. This is most important and this equation is also known as the equation for the hydraulic jump that is y 2 and y 1. For this these are the reference books. Thank you.