CE522 ENGINEERING HYDROLOGY

Jorge A. Ramirez

Level Pool Routing - Example

Use the Storage-Indication Method to route the Input hydrograph tabulated below.

This hydrograph flows into a reservoir whose storage and discharge characteristics are as presented in the following table. The initial storage in the system is 1'000,000 m³, and the initial outflow is 20 m³/s.

Reservoir or level pool routing refers to routing for systems whose storage and outflow are related by a function of the type S(t) = f[O(t)] which is of the invariable type (unique, non-hysteretic). These relationships imply that for a given set of conditions (e.g. stage) the outflow is unique, independent of how that stage is achieved. Reservoirs or systems with horizontal water surfaces have S vs. O relationships of the invariable type. Such systems have a pool that is wide and deep compared to its length in the direction of flow, and low flow velocities in the reservoir. For such systems, the peak outflow occurs when the outflow hydrograph intersects the inflow hydrograph.

The Storage-Indication method is a level pool routing procedure for calculating the outflow hydrograph of a system with horizontal water surface, given its inflow hydrograph, and storage outflow characteristics. The solution involves integrating the continuity equation as indicated below, and rearranging terms such that all the unknown quantities are on the left-hand side of the equation.

Storage-Indication Routing Equation:

For a level pool reservoir, the storage is a unique function of elevation; and the outflow is a unique function of elevation. Thus, the left-hand side of the equation above is a unique function of elevation in the system, only. Usually, the storage-elevation relationship is available from topographic surveys, and the outflow-elevation relationship is available from hydraulic considerations with respect to the outlet structures (e.g. spillways, etc.)

The solution involves the development of the function 2S/Dt + O = f(O) and then solving it sequentially for every time step. These steps are illustrated below.

1. Develop the function 2S/Dt + O vs. O. Use a Dt of 6 hours, as suggested by the time interval of the inflow hydrograph.

In the table above, Columns 1-3 are given. Columns 2 and 5 correspond to the desired function, 2S/Dt + O vs. O , which has been graphed above.

B - Proceed with the routing of the inflow hydrograph by using the Storage-Indication routing equation sequentially for every time step:

t = 0 -- i = 0. Initial Conditions: S_o = 1'000,000 m³; O_o = 20 m³/s.

t = 6 -- i = 1

(I_o + I₁) = (0 + 50) m³/s = 50 m³/s

(2S_o /Dt - O_o) = (2 x 1'000,000 m³)/(6 x 3600 s) + 20 m³/s = 72.593 m³/s

(2S₁ /Dt + O₁) = (I_o + I₁) + (2S_o /Dt - O_o) = 122.593 m³/s

Using the relationship (2S/Dt + O) vs. O developed in Part A, obtain the outflow O₁ corresponding to the value of (2S₁ /Dt + O₁) obtained above. This is done by entering the graph with the value of (2S₁ /Dt + O₁) and exiting with the value of O₁. Use interpolation as indicated below.

O₁ = 20 m³/s + [(34 - 20)/(190.4815 - 112.5925)] (122.593 - 112.5925) m³/s = 21.797 m³/s

t = 12 -- i = 2

(I₁ + I₂) = (50 + 130) m³/s = 180 m³/s

(2S₁ /Dt - O₁) = (2S₁ /Dt + O₁) - 2 x O₁ = 122.593 m3/s - 2 x 21.797 m3/s = 78.998 m3/s

(2S₂ /Dt + O₂) = (I₁ + I₂) + (2S₁ /Dt - O₁) = 258.998m³/s

Using the relationship (2S/Dt + O) vs. O developed in Part A, obtain the outflow O₂ corresponding to the value of (2S₂ /Dt + O₂) obtained above. This is done by entering the graph with the value of (2S₂ /Dt + O₂) and exiting with the value of O₂. Use interpolation as indicated below.

O₂ = 34 m³/s + [(57 - 34)/(320.8889 - 190.4815)] (258.998 - 190.4815) m³/s = 46.084 m³/s

Proceed as above for every time step. Results are tabulated below.

Time (h)	I (m3/s)	Ii + Ii+1 (m3/s)	2Si/Dt - OI (m3/s)	2SI+1/Dt + Oi+1 (m3/s)	O (m3/s)
0	0				20
6	50	50	72.593	122.593	21.797
12	130	180	78.998	258.998	46.084
18	250	380	166.829	546.829	97.129
24	350	600	352.572	952.572	168.889
30	540	890	614.794	1504.794	267.142
36	735	1275	970.509	2245.509	398.933
42	1215	1950	1447.644	3397.644	603.621
48	1800	3015	2190.403	5205.403	924.556
54	1400	3200	3356.291	6556.291	1164.37
60	1050	2450	4227.552	6677.552	1185.896
66	900	1950	4305.76	6255.76	1111.018
72	740	1640	4033.723	5673.724	1007.694
78	620	1360	3658.336	5018.336	891.347
84	510	1130	3235.641	4365.641	775.479
90	420	930	2814.684	3744.684	665.234
96	320	740	2414.216	3154.216	560.402
102	270	590	2033.411	2623.411	466.164
108	200	470	1691.084	2161.084	383.912
114	150	350	1393.261	1743.261	309.571
120	100	250	1124.119	1374.119	243.892
126	72	172	886.334	1058.334	187.707
132	45	117	682.921	799.921	141.863
138	25	70	516.195	586.195	104.087
144	10	35	378.022	413.022	73.366
150	0	10	266.291	276.291	49.134
156	0	0	178.022	178.022	31.761
162	0	0	114.501	114.501	20.343