Question

A stormwater detention basin is estimated to have the following storage characteristics Stage (m) Storage (m) 5.0 0 5.5 694 6.0 1525 6.5 2507 7.0 3652 7.5 4973 8.0 6484 The discharge weir from the detention basin has a crest elevation of 5.5 m and the weir discharge, Q, is given by: Q - 1.83H/2 Where H is the height of the water surface above the crest of the weir. T/eh catchment runoff hydrograph is given by: Time (h) 30 6090120 150 180 210 240 270 300 330 360 390 Runoff 0 2.4 5.6 3.4 2.8 2.42.21.81.5 1.2 1.0 0.56 0.34 0 m3/s) If the prestorm stage in the detention basin is 5.0 m, estimate the discharge hydrograph from the detention basin

this is a routing question

Answer 1

The pre-storm stage in the detention basin is 5.0 m. Hence, discharge occurs when water rises above the crest level. The flow rate at the stage height of 5 m is Omº/s. 3 Calculate the discharge at the stage height of 5.5 m by using the following equation: Q=1.83h ...... (1) Here, Q is the flow rate and h is the stage height. Calculate the stage height at 5.5m. h=5.5-5 = 0.5 m Calculate the discharge at the stage height of 5.5 m by using equation (1). Q=1.83 Q=1.83x(0.5) = 0.647 m/s Similarly, calculate the discharge for the remaining stage heights. Sample calculation: Calculate the value of (2S/At)+Q), corresponding to the stage height of 5.5 m. Here, S is the storage in the reservoir and At is the time interval. Substitute 1,041 mº for S, 30 min for At, and 1.16m/sec for Q. 250_2x(694)_0 647 30 — +Q = _ 2x(694) = -+0.647 30x60 =0.771+0.647 =1.418 m/s Similarly, calculate for the height of the water above the spillway crest. Tabulate the discharge-storage relationships used to determine the routing relationship as in Table (1). Storage, Stage Height Discharge, Q S (m /sec) |(m/sec) 5 5.5 694 1525 2507 3652 4973 6484 0 0 0.647 1.83 3.362 5.176 7.233 0.771111111 2.341444444 4.615555556 7.419777778 10.70155556 14.43744444 7.5 8 Plot a graph for the discharge, Q along the horizontal axis and the storage-outflow function [(2S/At)+Q) along the vertical axis by using the data from Table (1). From table (1): Draw the graph between storage-outflow function Q versus [(2S/At)+Q) as shown in figure (1) 1 2 3 4 5 6 7 8

Show the hydraulic routing equation (continuity equation). (1, +2)+(2- 2) = (2+0) ...... (2) Here, I is the inflow, Q is the flow rate, and S is the storage, At is the time interval, and suffix 1 and 2 denotes the start and end of considered time interval, At. Show the procedure to route inflow hydrograph. • Substitute the value of 11, 12, and (2S/ At)-Q in Equation (1) to get the value of (2S2 / At)+22 • Obtain the value of discharge Q2, at the end of considered time interval At, from the figure (1), corresponding to the value of (2S / At)+Q2- • Subtract 2Q, from (2S, / At)+Q2, to get the value of (2S, / At)-22. • Repeat the steps until the entire outflow hydrograph is calculated. Consider the time interval of first 30 minutes. The inflow at the start of first 30 minutes (t = 0min), I, is 0m/sec. The inflow at the end of first 30 minutes (t= 30 min), I, is 2.4mº/sec. Calculate the value of [(2S,/At)-2), from the reservoir properties. Here, S, is the storage in the reservoir corresponding to the time of 0 minutes and Q is the discharge from the reservoir corresponding to time of 0 minutes. Substitute Om for S,, Omº/s for Q,, and 30 min for At. 25,- 9,= 2x0m +om 30 min Calculate [(2S,/Af)+] by using equation (2). Substitute Omº/s for[(2S,/A1)-9], 0m²/sec for Iį, and 2.4 mº/sec for Iz. = (0+2.4)+0 = 2.4 m/s Obtain the value of outflow, Q, at the end of considered duration of first 30 minutes from figure (1) corresponding to the value of [(2S2/At)+Q2]. The value of outflow, Q2, corresponding to the [(2S2/At)+22] value of 2.4 mº/s is 0.68 m/s. Alternatively, method of Iteration can also be used to determine the value of outflow, Q2 form the reservoir properties in Table (1). Calculate the value of [(2S, / At)-2]. Substitute 3.6m°/s for [(2S2/41)+22] and 1.94m°/s for Q 2.4 – 2x(0.68 ) =1.04 m S Repeat the procedure for the time interval of next 30 minutes using the calculated values of [(2S2 / At)-22],[(2S2/At)+Q), and Q2. Use the calculate value of outflow at the end of first 30 minutes time interval as the inflow at the start of next 30 minutes time interval. Repeat the steps until the entire outflow hydrograph is calculated.

Tabulate the routed discharge hydrograph of the detention basin as in Table (2). Stage (m) (m/sec) m?sec) 2+971 m/sec) Il -Q 0 O 0 12.4 15.6 90 3.4 120 2.8 150 12.4 180 2.2 210 1.8 240 1.5 270 1.2 300 11 330 0.56 360 0.34 390 0 420 0 4500 480 0 0 2.4 9.04 9.52 6.66 598 5.44 4.88 4.24 3.68 3.2 1.04 0.52 0.46 0.78 0.84 0.94 0.98 0 0.68 4.26 4.53 2.94 2.57 2.28 1.97 1.63 1.34 1.09 0.77 0.48 0.23 0.04 0 0 2.58 1.94 1.32 0.86 0.78 0.78 1.02 1.04 0.98 0.86 0.78 0.78 0.78 0.78 | The final outflow hydrograph is tabulated in the column 4 of the table above.