 working as an assistant professor in a civil engineering department of Valgianist of Technology, Solapur. In the last lecture, we have seen what is the graduated flow and the dynamic equation for the graduated flow. Depending on that, in this session, we are going to solve some problems. So at the end of this session, you are able to find the most important thing that is the slope and the length of the channel which may be on the upstream or which may be on the downstream by using gradually ready flow equation and the step method. The first example, a rectangular channel of width 10 meter and normal depth of flow is 1.75 meter, has a bed slope 1 in 1600, obstruction in the form of overflow dam at the section rises the level of water by 1 meter. Calculate the slope of water surface with respect to horizontal line. Assume the Manning's coefficient as 0.025. Read the problem once again. Basically, we must know what is the field depth, what is the critical depth and what is the normal depth. Depending on this depth, three depths, we are finding out which type of slope it is. If yn is greater than yc, yn or we can say y0 also, greater than yc. That means normal depth is greater than critical depth. It is a mild slope when yn is or y0 is equal to yc, normal depth is equal to critical depth. It is a critical slope. When critical depth yc is greater than y0 or yn, which is a steep slope, then the fourth one is horizontal slope where the y is imaginary and adverse where y we cannot find out. Here depending on these slopes and three zones, 15 water surface profiles are expected. But as you know, C2 or critical slope 2 is not possible because yn and yc are coinciding. And in horizontal and adverse, zone 1 is imaginary. That's why these h1 and a1 are not possible. So five slopes and three zones were expecting 15, but C2, h1 and a2 are not possible where having only 12 water surface profiles. Now concentrate on this. First of all find out discharge. Discharge is area into velocity. Here velocity we have to find out with the formula of Manning's because Manning's coefficient is given to us that is 0.025. For Manning's formula that is 1 upon n, r is to 2 third and s is to half. Here r is hydraulic mean death or hydraulic mean radius. This is nothing but area upon weighted perimeter. Here area is 10 into 1.75 meter upon perimeter. This is a 1.75 on one side, on the other side 1.75 and the horizontal is 10. So we will get 17.5 upon 13.5 as r. This is 17.5 upon 13.5 as r. By calculating by substituting this value we will get discharge as 20.81 meter cube per second. Small q, it is a discharge per meter width. So q upon b, width is 10, it is 20.81 upon 10. It is 2.081 meter cube per second per meter. We cannot say it is meter square per second. It is meter cube per second per meter always. Critical depth we know that is from the Froude's number we have calculated it, yc is equal to cube root of q square upon g. This is the small q that already the value we have calculated. Put this value we will get yc is equal to 0.76 meters. y0 which is a normal depth given in the problem 1.75 meter. So 1.75 meter and by obstruction the water level rises by 1 meter which is given in the problem. Obstruction in terms of a small wear, the water level rises by 1 meter. So 1.75 plus 1, this is the field width that is 2.75 meter. Now see 2.75 is y, y0 is 1.75 and yc is 0.76 meter. This is a mild slope because normal depth is greater than the critical depth and y is greater than these two values so that y is greater than y0 which is greater than yc I can say this is a m1 type of water surface profile. Now by using the equation of gradual wear rate flow with respect to Manning's formula dy by dx is equal to h0 into bracket 1 minus y0 upon y bracket raise to 10 by 3 upon 1 minus yc upon y bracket cube. So these values we get 5.031 into 10 to the power minus 4. But the answer is asked the slope with respect to horizontal. So slope with respect to horizontal is nothing but s0 minus dy by dx, s0 is 1 upon 1600 and dy by dx we have calculated it. So the answer for this it is 1 upon 1 to 190. Here we have calculated only dy by dx. Here we have used the equation of gradual wear rate flow where Manning's formula is inserted so that the equation becomes equation representing with the depths. So my questions are you have to write the Manning's and Chej's formula and what is the criteria for mild surface profile? Chej's and Manning's formula are for measurement of velocity that is for Manning's it is v is equal to 1 upon n rs to 2 third and ss to half where n is Manning's coefficient r is hydraulic mean depth and s is a slope. Chej's formula is v is equal to c under root rs, c is Chej's constant r is hydraulic mean depth and s is a slope. And our question number 2 the answer is normal depth of flow is more than critical depth then it is a mild surface profile. A very wide rectangular channel carries a discharge of 3132 liters per second per meter width at a depth of 2 meter. A wear constructed across this channel rises the water level by 2 meter just upstream of it. Locate the point upstream of the wear at which the depth of water will be 3 meters. Assume bed slope of the channel as 1 in 4900. Assume value of c here Chej's formula is used as 50 use step method and they asked us to go for 2 steps. Here small q is directly given to us that is 3132 liters per second per meter but it is in liter can we are converting it into meter cube divide this equation by 1000 q is equal to 3.132 meter cube per second per meter y0 is given to us 2 meter. And now as we know yc that is calculated from the Froude's number yc is equal to cube root of q square upon g we will get yc as 1 meter. Now depth of water just upstream of it it is rises by 2 meter y0 plus 2 it is 4 meter that is y so y is 4 meter y0 is 2 meter and yc is 1 meter. So y is greater than y0 which is greater than 1 I told you previous also in the previous example also the same situation was there 4 is greater than 2 is greater than 1 which is a m1 type of profile and mild slope. Now what is the specific energy? Specific energy is nothing but the y or the you can say the p by w or pressure head plus velocity head here we are not going to take the static head or datum head we are not going to take. Specific energy is nothing but it is a addition of p by w that is y and kinetic head we score upon or we score upon 2g we can say put these values find out es1 similarly find out es2 here es2 is y2 plus v2 square upon 2g similarly find out es3 so es1 es2 es3 we have calculated as 4.03 3.54 and 3.055 meters for a very wide rectangular channel already we have defined the r hydraulic mean depth is taken as y so in the Manning's formula wherever there is a r we have to keep y that is for wide rectangular channel not for any other channel otherwise we have to calculate separately a by p put these values put these values we will get sf which is 6.13 similarly sf2 similarly sf3 then we have to find out the averages of this through sf12 sf23 put these values delta l1 is nothing but es2 minus es1 upon s0 minus sf mean sf mean of 1 2 first and then 2 3 second in the 1 2 we are getting 3 at 4 0 and the second we are getting 5 to 6 0 take a addition of these 2 it is 9.1 kilometers this is the perfect procedure for solving by step method or we can say direct step method we can solve this by single step also in the next problem we will compare this method with a single step directly and we will compare the results these are some reference books for you thank you