Storm Drainage Design Project Study Essay Example | Topics and Well Written Essays - 2000 words

Storm Drainage Design Project Study - Essay ExampleAs we go along this stock work ,we will be able to understand what hydrographs be.The graphs above are the results of the Cynon river study data. The study took 4 days to finish nonstop. The measurements of the river height and the elicit are done every hour for 96 hours. The rainfall was plotted using the bar graph and the discharge was plotted using the line graph. In the psychoanalysis of the rainfall, you will notice that the rainfall is fluctuating. It is not as though on that point is a steady rise in the rainfall. The line graph shows the rise of discharge of water in the river, As the rainfall increases, the discharge also increases. The graph satisfy the components of a hydrograph. From the start of the study, you will notice that there is almost a steady flow of water in the river. That means that there is no increase in rainfall. At the start of the 44th hour, the water start to rise. This bit of the graph is called the rising arm. This is the part of a hydrograph when water rises too the point of peak discharge. After it reached the peak point, the water stars to recede and this part is called the falling limb or the receding limb. This part denotes that rainfall is finally over and that the accumulated water in the river starts to stabilize again. The part of a hydrograph that is the highest point is called the peak discharge.. this is when there is the greatest amount of water in the river. The lag time is the period of time taking place between the peak rainfall and peak discharge.ComputationsBy the application of the Mannings Formula, we will be able to get the pry of breadth b of the open channel with the following dataChannel design attached Data Q = 1.0 m3/s n =0.012S= 1/2000 = 0.0005d= 0.5Formula to be used V= where v = velocityQ= AvR=hydraulic RadiusQ= AS=slopeA= bdn=Mannings coefficientR= Q=dischargeComputationsA= db= 0.5(b)Q= AR= 1.0 = 0.5b 1.(0.012) = 0.5b 0.012 = 0.5b = 0.50.54 29 = 0.5 = 1.0858 = (1.0858)3 = b31.2801 = 1.2801 = 1.2801 (1.0 + 2b + b2) = 0.25b51.2801 + 2.5602b + 1.2801b2 = 0.25b51.2801 + 2.5602b + 1.2801b2 - 0.25b5 = 0 b = 2.2104 m. The value of depth of the river is also needed in order to solve for the value of the discharge of water in the river. The acquired value for depth will help us acquire the value fro the cross-section(a) area of the river. In that way, we will be able to solve for the value of the discharge on the river. ComputationsQ=AvwhereA=cross-sectional areav=velocity = 4.0 m/sA=bdb=15 m.A=15(d) R = R = v = v =4.0 = 4.0(0.012) ==2.1719 =(2.1719)3 =10.2451 =10.2451(225 + 60d + 4d2) = 225d2 2,305.1475+ 614.7069d + 40.9804d2 = 225d2 2,305.1475+ 614.7069d + 40.9804d2 - 225d2 = 02,305.1475 + 614.7069d + 1 84.0196d2 = 0By quadratic equation solve for the