Route the inflow hydrograph given below
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- 8.1 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.Q1: given a reservoir with a storage discharge linear relationship governed by the equation: S-KQ, if the K-8000 and the initial storage is 30000m³,rout the inflow hydrograph given below using At=1hr. S 0 1 2 3 4 5 6 7 8 Time, hr. 100 200 0 400 300 200 100 50 Inflow m³/sec 0(b) A simple reservoir has a linear storage indication curve defined by Q/2 = S/At, where At = 1.0 hr. If S at 8 A.M. is zero ft³/s-hr, use the continuity equation to route the following hydrograph through the reservoir. Time (hr) 9 A. M. 11 A. M. 12 Noon 120 Inflow (ft³/s) 65 110 8 A.M. 40 10 A. M. 80 1 P.M. 170
- Q1. Find linear relation between Discharge (Y cumec) and Stage (x m) of for the below hydrologic data: X Y X Y X Y X Y 18 905.7 19 1004.66 1 12.86 19 1004.66 17 811.9 12 420.3 19 1004.66 88.78 12 420.3 3 40.5 8 199.9 723.26 5 88.78 17 811.9 3 40.5 639.78 4 62.06 6 120.66 7 157.7 199.9 5 88.78 13 488.3 10 299.78 1 12.86 488.3 5 88.78 12 420.3 14 561.46 811.9 12 420.3 17 811.9 14 561.46 157.7 3 40.5 3 40.5 10 299.78 88.78 811.9 6 120.66 1 12.86 1108.78 6 120.66 10 299.78 1108.78 157.7 905.7 16 723.26 561.46 12.86 3 40.5 62.06 1004.66 9 247.26 88.78 639.78 488.3 639.78 120.66 811.9 12 420.3 357.46 7 157.7 13 17 7 5 20 18 1 19 17 11 1215256 17 20 7 13 LASZO 14 4 5 16 15 8 5 17 811.9 157.7 7Q2: For a certain river, the following data were obtained for the cross section at a selected station: Station 0 1 2 3 4 5 6 X(m) 14 18 21 Y(m) 3 0 Velocity at 0.6 of the depth (m/sec) 0.4 0 0 0 3 6 2 0.5 0.7 4 10 6 0.6 For this cross section, estimate the discharge in cubic meter per second 5 0.5 0(a) Route the inflow hydrograph given below through a reservoir. The storage data (surface elevation versus storage volume) for the reservoir are also given below. The spillway discharge is expressed by the equation (Q = 3 LH2). The crest height of the spillway is 50 ft. and the length of the spillway is 35 ft. Inflow Hydrograph: Time (days) Inflow (ft/s) 0 70 185 360 0 0.5 1 1.5 |2 2.5 3.0 3.5 4.0 480 300 165 80 0 Storage Data: Elevation (ft.) Storage (acre-ft.) 50.0 50.5 51.0 51.5 52.0 52.5 53.0 53.5 231 247 277 313 353 400 452 509 (b) A simple reservoir has a linear storage indication curve defined by Q/2 =S/At, where At 1.0 hr. If S at 8 A.M. is zero f/s-hr, use the continuity equation to route the following hydrograph through the reservoir. Time (hr) Inflow (ft'/s) 8 A. M. 10 A. M. 80 11 A. M. 1 P.M. 9 А. М. 12 Noon 40 65 110 120 170
- N S (b) The hydrograph resulting from an 8-hr storm over a catchment area of 4000 km² is given in the table below. Derive and plot the unit hydrograph for this storm. Time hrs Discharge (m³/s) 0 6 12 18 24 30 36 42 48 54 60 120 140 480 1980 2600 2500 1500 1100 540 450 400 Baseflow (m/s) 120 140 160 180 210 220 280 295 360 380 400 → 00 5 2NUse the level pool method to route the inflow hydrograph given below. The storage- outflow characteristics are also given below. Use a time interval (time step) of 10 minutes in your calculations. Initially the storage is zero. Compute the outflow, and plot the inflow and outflow hydrographs in the same figure. Make sure to perform routing for at least 6 hours. Storage (ft³) Discharge 0 32670 65340 98110 130680 163350 196020 0 5 12 25 30 38 54 (cfs) Storage (ft³) 228690 261360 294030 326700 Discharge (cfs) 90 100 120 140 Time (min) 0 10 20 30 40 50 60 70 80 90 Inflow (cfs) 0 15 25 35 35 35 40 55 62 60 Time (min) 100 110 120 130 140 150 160 170 180 190 Inflow (cfs) 50 42 35 30 25 20 15 10 0 0 Time (min) 200 210 220 230 240 200 Inflow (cfs) 0 0 0 0 0 0Question 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 700
- A reservoir has the elevation-discharge-storage relationships as shown Storage x 106 (m³) Elevation Outflow (m) 299.5 Discharge Q (m³/s) 5.4 300.2 6.1 15 300.7 6.6 28 301.2 7.2 46 301.7 7.8 81 302.2 8.5 121 302.7 9.4 166 Develop the relationships using At = 3 hours for: (i) Elevation vs S+ [(Q At) / 2] (ii) Elevation vs Outflow Discharge Q, by plotting them on the same graph раper.Q4-The stream flows due to three successive storms of 2.9, 4.9 and 3.9 cm of 6 hours duration each on a basin are given below. The area of the basin is 118.8 km². Derive a 6-hour unit hydrograph for the basin. An average storm loss of 0.15 cm/hr can be assumed. Time (hr) 0 Flow (cumec) 20 3 50 6 92 9 140 12 199 15 202 18 204 21 24 84.5 27 30 33 45.5 29 20The given table shows a unit hydrograph of 2-hour duration. Derive 4-hour unit hydrograph for the same catchment. And also draw a graph Time (hours) 0 2 4 6 8 10 12 Flow (m3/sec) 0 86 122 93 52 22 0