 Deshmukh Sachin working as an assistant professor in civil engineering department of wall chain stop technology. The last lecture we have seen the dynamic equation for rapidly varied flow as well as the equation for the hydraulic jump which is the most important example of rapidly varied flow. In the today's lecture we are going to solve problems on rapidly varied flow. After solving you are able to know or you can solve it carefully the problems on hydraulic jump. You can classify the hydraulic jump on the basis of the Froude's number. I repeat once again Froude's number is very important in the case of hydraulic jump. It is the ratio of inertia force to the gravitational force and you can also calculate the loss of energy through hydraulic jump. The equation for the energy loss through jump that also we have seen in the last lecture. Let us see this problem. Calculate the height of hydraulic jump from the following data. The data is given as discharge through the channel that is 6 meter cube per second the depth of water before the jump that is y1 that is 1 meter is given width of the channel is also given that is B is given as 1 meter. First of all read the problem twice and then write down the data given in the problem. Here discharge is given as 6 meter cube per second denoted as capital Q. Death before jump y1 is 1 meter width B is 1 meter. Now death after the jump y2 is calculated from the following equation that equation also we can call the equation for the conjugate depth that is y2 upon y1 is equal to half into bracket minus 1 plus under root of 1 plus 8 fr1 square where y2 and y1 are the depth after and before the jump frouds fr is the frouds number before the jump 1 is for before the jump 2 is for the after the jump. Now using this equation put the values we will get v1 upon g y1 is equal to v is Q into 1 meter area into divided by 1 meter that is y is 1 meter width is also 1 meter v is Q by A. So, we will get 6 that is discharge upon 1 into 1 is 1 divided by 9.81 into 1 that is v upon under root g y1 that is froude's number we are calculating. Froude's number we have to calculate to know the type of jump. Froude's number is v upon under root g y1 and from that we have calculated as 1.915. So we can conclude here jump is possible first of all because the criteria is that whenever the froude's number is more than 1 jump is going to be created. Using this froude's number find out y2. So y2 is equal to 1 half into bracket same equation under root 1 plus 8 fr1 square minus 1 and y1 is 1 that is why it is taken as directly on the right hand side. So you putting these all values we will get 2.254 meters. So y2 is 2.254 meters the depth of the jump is calculated from y2 minus y1. So 2.254 minus 1 it is 1.254 is the depth of hydraulic jump. So we can write as for the froude's number is 1.915 the jump is weak jump because the classification is 1 to 1.7 which is a unduler 1.7 to 2.5 it is a weak 2.5 to 4.5 it is oscillating 4.5 to 9 it is a steady and above 9 it is a strong jump. Now these are the questions write the equation used to measure the conjugate depths conjugate depths that is y2 and y1 or you can say sequent depths also. And second is classify the hydraulic jumps on the basis of froude's number answer for the question number 1 y2 upon y1 is equal to 1 half into bracket minus 1 plus under root of 1 plus 8 fr1 square and for the answer for the question number 2 as per the froude's number 1 to 1.7 unduler 1.7 to 2.5 weak 2.5 to 4.5 oscillating 4.5 to 9 steady and above 9 strong jump. Now we see the second problem a discharge of 840 meter cube per second flows down a spillway and then passes on a 60 meter wide concrete apron in bracket n is given to us that is 0.014 the velocity of the water at the toe of spillway is 10 meter per second a tail water depth of 4.4 meter the channel below causes a hydraulic jump on the horizontal apron. Determine depth before the jump, length of the jump, energy lost in the jump and specific force at the toe is a very important problem. Read the problem once again and write down the data given in the problem. Here particularly in this problem n is given to us n, n is the Manning's constant. That means we are going to use the Manning's formula particularly in this equation to find the velocity. The Manning's formula is v is equal to 1 upon m r raise to 2 third s raise to half r is hydraulic mean depth and s is a channel bottom slope. So data given q is equal to 840 meter cube per second width is 60 meter n is 0.014 velocity 1 that is 10 meter per second and y2 y2 is given to us 4.4 meters we have to find out here y1 now. Now using the relation 2 q square upon g is equal to y1 y2 into bracket y1 plus y2 I told you in the before also we have already we have served this equation when we are starting the momentum equation to derive the equation for the conjugate depth. So 2 p square upon g is equal to y1 y2 into bracket y1 plus y2. Now using this equation get a quadratic equation y1 square plus 4.4 y1 minus 9.018. Solving for y1 we will get 1.531 meters. So length of the jump so y1 we will get now we will calculate the length of the jump as per usbr united states of bureau of reclamation that is usbr has given the equation that is length of the jump is equal to 526.9 into bracket y2 minus y1 here the constant we are taking as 6.9 that for the extreme case so 6.9 into bracket 4.4 minus 1.531 y2 minus y1. So the length of the jump we got that is 19.796 meters. Now we will calculate the loss of energy through the jump. We know the equation y2 minus y1 bracket cube upon 4 y1 y2 already we have derived it. Put the values of y1 y2 we get loss of energy as 0.876 meters. Now the important here that is specific force Fs that we have to calculate Fs is area into y bar plus q square upon a g. Put these values into y bar plus q square in this equation a y bar plus q square upon a g. A is width into depth b into y1 square because y bar is y1 upon 2 so b y1 square upon 2 plus q square as it is upon area is b y1 into g. Put the values 60 q square into 1.531 y1 plus q square plus q square upon 2 plus 840 square upon 9.81 into 1.531 into 60. 60 is a width 60 is a width. So you will get 853.32 newtons at the top of the spillway. So these are the answers of this problem for this particular topic. You can refer these books. Thank you.