 Good morning students, myself, Deshmukh Sachin, working as an assistant professor in the Civil Engineering Department of Quality and Stock Technology. Today we are going to learn the concept of energy loss through hydraulic jump. Before this, we have learned about gradually varied flow and rapidly varied flow. And you are knowing hydraulic jump is the best example of rapidly varied flow. At the end of this topic, you are able to explain what is the concept of loss of energy through hydraulic jump. You are able to derive the equation of energy loss through the hydraulic jump. And you are able to solve the problems on loss through hydraulic jump. And also you can classify the hydraulic jump on the basis of Froude's number. Froude's number is very important as the gravitational force is predominant than the other forces that is viscous, then pressure, elastic and surface tension. Particularly in this hydraulic jump phenomenon, this gravitational force is predominant and that is why the Froude's number is taken into account. Let us see how the energy loss takes place. When the hydraulic jump is formed, the formation of eddies and turbulence takes place resulting in the loss of energy. If this loss of energy per Newton of liquid is denoted by delta E, then applying Bernoulli's energy theorem to the section 1, 1 and 2, 2 in which the hydraulic jump is formed. This is section 1, 1 and this is section 2, 2. This is section 1, 1 and this is section 2, 2. This is the velocity distribution, here also velocity distribution. Now just see this section once again carefully. y1 plus u1 square upon 2g is equal to y2 plus v2 square upon 2g plus loss that is delta E. If you can further simplify, you will get delta E is equal to q square upon 2g y1 square minus q square upon 2g y2 square minus into bracket y2 minus y1. Here q we got from v1 into y1, this q is nothing but a discharge per meter width. That is why area is y1 into 1, therefore q is equal to v1 into y1 and v2 into y2 as a continuity equation. Further simplifying, we will get the equation delta E is equal to 2q square upon 2g 2q square upon g y1 y2 square into bracket y2 square minus y1 square upon 4 y1 y2. minus into bracket y2 minus y1. Now applying momentum equation to the hydraulic sum that is the net force is equal to rate of change of momentum. That is p2 minus p1 that is the net force is equal to rate of change of momentum that is rho that is mass density into q into bracket change in the velocity. As the net force is acting from right hand side to left hand side as we have seen in the figure, the change in the velocity should be taken in the same direction v1 and v2. Put these values here and simplify it, we get q square upon g into bracket 1 minus y1 minus 1 upon y2. Again simplifying, we will get this is a very important equation 2q square upon g y1 y2 is equal to y2 plus y1. This is a very important equation we got and we are going to insert this particular equation in the main equation of hydraulic jump. Just remember this equation once again 2q square upon g y1 y2 is equal to y2 plus y1. You can write this equation on the top of your page also. Substituting this value in the above equation for loss of energy, we will get delta e is equal to y2 minus y1 bracket cube upon 4 y1 y2. This is the equation used to calculate the energy loss through hydraulic jump. You have to simplify this equation slowly and you can calculate delta e is equal to y2 minus y1 bracket cube upon 4 y1 y2. The basic equation which we are using for this hydraulic jump is a momentum equation not an energy equation. Considerable loss of energy takes place in a very small length. That is why we are taking the momentum equation and we have calculated that loss of energy is equal to y2 minus y1 bracket cube upon 4 y1 y2. Also we can say it is a specific energy at section 2 minus specific energy at section 1. y2 minus y1 is nothing but the height of the jump. So height of the jump cube upon 4 y1 y2. Relative loss of energy is the ratio of loss of energy upon specific energy at 1. And efficiency of the jump is the ratio of specific energy after jump to the specific energy before jump. And which we can write efficiency that is es2 upon es1 es2 it is specific energy that is y2 plus v2 square upon 2g upon y1 plus v1 square upon 2g. For deriving this equation following the assumptions the channel bottom is practically horizontal. It is the channel bottom must be smooth. Channel bottom is prismatic that is no scoring no silting and hydrostatic pressure distribution is valid. And the entire flow is a stream line that is in the straight line. This is the question for you. Write the equation used to calculate energy loss through hydraulic jump. This is the equation delta e is equal to y2 minus y1 bracket cube upon 4 y1 y2. Now we will see what are the types of jumps. The types of jumps are classified on the basis of Froude's number. The United States Bureau of Reclamation commonly called as USBR have classified these jumps on the basis of Froude's number. That is 1 to 1.7 it is a unduler jump Froude's number 1 to 1.7 and 1.7 to 2.5 it is a weak jump. 2.5 to 4.5 it is a oscillating 4.5 to 9 it is a steady and above 9 it is a strong jump. This type of jump that strong jump we can observe on a field. Length of the jump we can calculate as per the USBR. The length of the jump is defined as the horizontal distance from beginning of the jump to the point of subcritical region of the flow where the water surface having maximum depth becomes horizontal. As per USBR length of the jump is equal to k into bracket y2 minus y1 where k is obtained by experimental results varies from 5 to 6.9. The location of the jump these are some locations. The exact location of the jump is very important to know or to design for the spillway. Formation of hydraulic jump on the downstream side of the Swiss gate. Formation of the hydraulic jump in the channel having break in its bottom slope and formation of hydraulic jump behind overflow of the barrier. The hydraulic jump may also occur at the top of the spillway that is at the end of the spillway and downstream side of the hump that is a elevated portion. Lastly these are some of the uses. The hydraulic jump serves as an efficient means of dissipating excess energy. The hydraulic jump is used to increase the level of water in the downstream of the flumes measurement. That is in the flumes the area is contracted so the height is increased. The hydraulic jump is also used to reduce the uplift pressure by increasing the effective width of apron. It serves to intimate mix the chemicals that is in water purification. It is used to area the stream water that is polluted and it serves as an aid in intense mixing and gas transfer in certain chemical process. And lastly it is used for removal of air pockets from the water supply lines thus preventing air logging. So these are some of the important uses of the hydraulic jump. These are the reference books for you. Just take the name of the reference books. Thank you.