One tributary drains an area of 3.6 ha which was planned to develop for high-density residential area with 95% fraction imperviousness.

Assignment Task: Task: Project Brief An urban catchment located in Melbourne, Victoria, which has a total area of 9.00 ha, is composed of two main tributaries shown in Figure 1. One tributary drains an area of 3.6 ha which was planned to develop for high-density residential area with 95% fraction imperviousness. The remaining part of the catchment has impervious surfaces covering 1.0 ha. It has been estimated that it only takes 5 minutes for rainwater to travel from the furthest point in the high-density residential part of the catchment to the catchment outlet. An 80.0 m long stormwater drainage pipeline is proposed to service the area as shown in the Figure 1. The new line is to be connected to an existing system at an existing downstream pit. A minimum 1200 mm diameter pipe is specified for maintenance purposes, and the asset owner requires a maximum velocity of 6.0 m/s and design life of 100 years. It was required that the total energy head of the pit to be 0.50 m below the finished surface level. Consider pit loss coefficient 0.50. The existing system discharges the water into a small trapezoidal channel having 0.65% bed slope, 1:1 (H:V) side slopes and Manning’s roughness n = 0.035 as shown in Figure 1. A pipe culvert is proposed across the river. At the top of the culvert, there is a roadway of two lanes (3.50 m lane width and 1.5 m shoulder width). The crown level of the road is 125.65 m with 3% road crossfall towards shoulders from centerline. The allowable flow velocity along the channel is 2.5 m/s. Thickness of the road pavement is 460 mm and it is to be the minimum cover over the pipe. Inlet and outlet invert levels (IL) of the pipe are 123.85 m and 123.75 m respectively. Required freeboard below shoulder level is 100 m Note: 1. For 1% AEP event, runoff coefficient values for the impervious and pervious portions of the sub-catchments can be considered as 0.90 and 0.40, respectively. Use the general formula of calculating equivalent “CA” from non-homogenous catchments, i.e. (CA)eqv. = C1 × A1 + C2 × A2. 2. While perform activities, you must follow WHS/OHS requirements. 3. Consult with your instructor for the safety, accuracy, and performance. 4. You must follow relevant Australian standards in design calculation. 5. You must use appropriate design aids. Your Tasks: ? Calculate time of concentration (tc) for the second sub-catchment having 1.0 ha impervious using Pilgrim Equation. ? Download IFD table from Bureau of Meteorology (BoM) and attach. ? Determine the peak runoff from the catchment for a 1% AEP event. ? For DN1200 pipe, calculate the difference in water surface elevation between the two pits. ? Determine the shoulder level of the road. ? Calculate the allowable headwater depth. ? Determine the suitable pipe size (number and diameter) for the culvert. ? Calculate the design discharge for each of the pipes. ? Determine the inlet control headwater depth. ? Determine the outlet control headwater depth. ? Determine the outlet flow velocity. This VU22547 : Engineering Assignment has been solved by our Engineering Experts at TVAssignmentHelp. Our Assignment Writing Experts are efficient to provide a fresh solution to this question. We are serving more than 10000+ Students in Australia, UK & US by helping them to score HD in their academics. Our Experts are well trained to follow all marking rubrics & referencing style. Be it a used or new solution, the quality of the work submitted by our assignment experts remains unhampered. You may continue to expect the same or even better quality with the used and new assignment solution files respectively. There’s one thing to be noticed that you could choose one between the two and acquire an HD either way. You could choose a new assignment solution file to get yourself an exclusive, plagiarism (with free Turnitin file), expert quality assignment or order an old solution file that was considered worthy of the highest distinction.