Question

Consider a 2-acre detention basin with vertical walls. The triangular inflow hydrograph increases linearly from zero to a peak of 60 cfs at 60 min and then decreases linearly to a zero discharge at 180 min. Route the inflow hydrograph through the detention basin using the head discharge curve for the 5-ft pipe spillway in table below. The pipe is located at the bottom of the basin. Assuming the basin is initially empty, use the level pool routing procedure with 10-min time interval to determine the maximum depth in the detention basin.

Elevation Discharge

(ft) (cfs)

0.0    0

0.5    3

1.0    8

1.5 17

2.0 30

2.5 43

3.0 60

3.5 78

4.0 97

4.5 117

5.0 137

5.5 156

6.0 173

6.5 190

7.0 205

7.5 218

8.0 231

8.5 242

9.0 253

9.5 264

10.0 275

Answer 1

Inflow values are interpolated and values were obtained for each 10 min interval.

Time (min)	Inflow (cfs)
0	0
10	10
20	20
30	30
40	40
50	50
60	60
70	55
80	50
90	45
100	40
110	35
120	30
130	25
140	20
150	15
160	10
170	5
180	0

H (ft)	Discharge Q (cfs)	Storage (ft3)	2S/t	2S/t + Q
0	0	0	0	0
0.5	3	43560	145.2	148.2
1	8	87120	290.4	298.4
1.5	17	130680	435.6	452.6
2	30	174240	580.8	610.8
2.5	43	217800	726	769
3	60	261360	871.2	931.2
3.5	78	304920	1016.4	1094.4
4	97	348480	1161.6	1258.6
4.5	117	392040	1306.8	1423.8
5	137	435600	1452	1589
5.5	156	479160	1597.2	1753.2
6	173	522720	1742.4	1915.4
6.5	190	566280	1887.6	2077.6
7	205	609840	2032.8	2237.8
7.5	218	653400	2178	2396
8	231	696960	2323.2	2554.2
8.5	242	740520	2468.4	2710.4
9	253	784080	2613.6	2866.6
9.5	264	827640	2758.8	3022.8
10	275	871200	2904	3179

S0 = 0 ft3/s; Q0 = ft3/s

2S1/t - Q1 = 2x0 -0 =0

2S2/t + Q2 = (I1+I2) + 2S1/t - Q1

Find Q1 by interpolating values from above table for corresponding 2S2/t + Q2 value

2S1/t - Q1 =  2S2/t + Q2 - 2Q1

Repeat this process

Time (min)	I (ft3/s)	Ii+Ii+1 (ft3/s)	2Si/t - Qi	2Si+1/t + Qi+1	Q (ft3/s)
0	0				0
10	10	10	0	10	0.2024
20	20	30	9.5952	39.5952	0.8015
30	30	50	37.9922	87.9922	1.7812
40	40	70	84.4298	154.4298	0.7074
50	50	90	153.015	243.015	3.6563
60	60	110	235.7024	345.7024	3.7608
70	55	115	338.1807	453.1807	1.5477
80	50	105	450.0853	555.0853	9.9217
90	45	95	535.2419	630.2419	3.5976
100	40	85	623.0466	708.0466	9.9912
110	35	75	688.0642	763.0642	14.5122
120	30	65	734.0397	799.0397	5.6484
130	25	55	787.7428	842.7428	10.2289
140	20	45	827.2849	872.2849	13.3252
150	15	35	845.6345	880.6345	14.2003
160	10	25	852.2339	877.2339	13.8439
170	5	15	849.5461	864.5461	12.5141
180	0	5	839.5179	844.5179	10.4149
190		0	823.6880	823.6880	8.2318
200		0	807.2244	807.2244	6.5063
210		0	794.2118	794.2118	5.1424
220		0	783.927	783.927	4.0645
230		0	775.7980	775.7980	3.2125

Maximum discharge = 14.5122 ft3/s. So depth of detention basin required = 1.5ft (corresponding to 17 ft3/s discharge from elevation discharge table). If exact value is required interpolation can be used.

So depth of basin = 1.33ft ~1.35 ft