Analyse the given water distribution network using Hardy Cross method. Data: Inflow = 500 m³/sec Outflows (Q1,Q2,Q3)= 166.666 m³/sec K=4 K=2 K=3 K-1 K=1 K-3 K-3 K=2 K=3 K=2 K-1 K-3 K=1 Inflow
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- 10-min unit hydrograph for a catchment is given in Table 1 below. Table 1: 10-min Unit Hydrograph Time (min) 05 10 15 20 25 30 35 40 45 50 1.75 1.95 2.0 1.8 1.25 0.6 Flow (m³/s) 0 0.5 1.0 1.4 0 a. What is the area of the catchment? b. Construct a storm hydrograph resulting from the 40-min rainfall event. The rainfall excess depths are provided in Table 2. Determine the peak discharge resulting from this 40-min rainfall event. Table 2: Rainfall Excess Depths Rainfall Excess (mm) Time (min) 0-20 4 20-40 6Using the following values, give the water budget equation for this lake, and solve for annual groundwater flow. There are no other components. Period of time: Calander Year.Lake volume at beginning of year: 10,000. m^3Lake volume at end of year: 9,000. m^3Stream inflow: 700. m^3/month ET: 54.0 m^3/monthGW flow: unknown Give appropriate sig figs and show work Budget Equation: Inflows = Outflows + Change in StorageOTO 'T Hydrology Flood Routing • HW • Q2// . • For a large catchment, the precipitation and runoff are being recorded monthly. The records for 2 years are tabulated in the following table: The variables are assumed to be linearly related. Work out a relationship between the monthly precipitation and runoff for the location and use the relationship to estimate the expected amount of runoff generated when monthly precipitation is 14 cm (1 ||| = t + فلاتر O Moretly precipitatio runoff ntemile temat Jan-10 Feb-10 Mar 10 Apr 19 May 19 Jun-10 19 Jul-10 Aug 10 ADET Sep 10 Oct-10 Nov-10 Dec-10 Jan-11 Feb-11 Mar-11 Apr-11 May 11 Jun-11 Jul-11 Aug-11 Sep-11 Oct-11 Nov-11 Dec-11 ===============mage 64 65 5.1 71 74 21 7.1 10.2 9.9 8.4 58 10.1 7.3 5.5 114 10.8 7.5 8.2 79 4.3 10.4 3.9 2.4 11 1.3 05 18 2 43 1 3 47 44 2.8 14 6.3 44 1.8 4.2 إضافة شرح... 0 0 5.9 × mum
- Question 2 Route the inflow hydrograph through a reservoir as indicated in Table 2. The crest height of the spillway is 50 ft and storage capacity at this level is 116 ft3/s-day. The reservoir routing curves graph is given in Figure 1. Perform calculation until the outflow value is equal to zero. Based on the obtained result, plot the inflow and outflow hydrographs. Table 2 0.4 1.6 2 2.4 2.8 Time (day) 0 Flow (ft³/s) 0 0.8 120 1.2 229 56 380 268 155 50 Figure 1 25/atO (ft/s.day)-primary x-axis 700 800 900 1000 54.0 53.5 53.0 52.5 52.0 51.5 Elevation vs Storage (curve a) Elevation vs 25/De-O (curve b) Elevation vs Outflow (curve c) 51.0 50.5 50.08 49.5 110 230 0 Water Surface Elevation (tt) 400 W 500 130 100 600 150 200 190 170 210 Storage (ft³/s.day) - secondary x-axis 300 400 500 Outflow (ft³/s.day) - tertiary x-axis 600 250 1100 7008.1 The storage, elevation and outflow data of a reservoir are given below: Outflow discharge (m³/s) Elevation (m) 299.50 300.20 300.70 301.20 301.70 302.20 302.70 Storage 10 m² 4.8 5.5 6.0 6.6 7.2 7.9 8.8 0 0 15 Time (h) 0 3 6 9 12 Discharge (m/s) 10 20 52 60 53 40 75 115 160 The spillway crest is at elevation 300.20 m. The following flood flow is expected into the reservoir. 15 18 21 24 27 43 32 22 16 10 If the reservoir surface is at elevation 300.00 m at the commencement of the inflow. route the flood to obtain (a) the outflow hydrograph and (b) the reservoir elevation vs time curve. 8.2 Solve Prob. 8.1 if the reservoir elevation at the start of the inflow hydrograph is at 301.50 m.A catchment is idealized as a 30km × 30km square. It has four rain gauges as . During a month, the precipitation at these gauges is measured as A=(309) mm, B=(292) mm, C=(279) mm and D=(297) mm. Compute the average precipitation (in mm, up to one decimal place) over the catchment during this month by using the Thiessen polygon method. [11]A BD C
- Given below are ordinates of hydrograph, separate the base flow from direct runoff flow by using the fixed base method, and then find the unit hydrograph (UH) with all graphics. Note: using basin area 8000 km² Time (day) 0 1 2 3 4 5 6 7 8 9 10 11 12 Flow (m³/s) 200 180 180 220 300 500 450 420 350 300 220 180 160A catchment is idealized as a 30km x 30km square. It has four rain gauges as shown in the Figure Q4. During a month, the precipitation at these gauges is measured as A=(300 mm, B=(285 ) mm, C=(272 ) mm and D=(290- ) mm. Compute the average precipitation (in mm, up to one decimal place) over the catchment during this month by using the Thiessen polygon method. A D Figure Q4- Watershed boundary and rainfall gaugesPerform the flood routing for a reach of river given X = 0.15 and K = 3 days using the inflow hydrograph with At = 1.5 day and At = 3 day below. Fill out the table and show detailed steps and calculations for each At for the first 3 rows. What is the effect of the magnitude of At on outflow hydrograph? Time (days) 0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18 19.5 21 Inflow (cfs) Outflow (cfs) 0 5 13 34 68 74 73 61 45 32 27 21 14 7 0 Time (days) 0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 Inflow (cfs) 0 13 34 68 74 73 61 45 32 27 21 14 7 0 Outflow (cfs)
- Please provide complete solution. 1. Neglecting the losses, find the discharge through the venture meter. a. 0.0314 m^3/s b. 0.0404 m^3/s c. 0.0507 m^3/s d. 0.0870 m^3/s 2. Considering head loss between point 1 and 2 equal to 20% of point 1 velocity head, Calculate the rate of flow. a. 0.0210 m^3/s b. 0.0576 m^3/s c. 0.0401 m^3/s d. 0.0311 m^3/sRoute the inflow hydrograph tabulated in the following table through a river reach for which x = 0.2 and K = 2 days for only the first three days. Use a routing period At=1 day and assume I = O for the first day. Time (day) 1 2 3 Inflow (cfs) 4000 7000 11000 :Derive and plot 3-hour Unit hydrograph for a stream which drains an area of around (401) km2. Time-discharge data of the storm is given is in Table. Assume the base flow as 2 m3 /sec. Time (Hrs.) 0 3 Flow (m³/s) 2 6 6 9 11 13 12 15 18 21 10 8 6 4 24 2