CE322 Basic Hydrology

Jorge A. Ramirez

Level Pool Routing - Example

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

Time

(h)

Input Hydrograph (m3/s)

Time

(h)

Input Hydrograph (m3/s)

0

0 0.00

90

420.000

6

50.000

96

320.000

12

130.000

102

270.000

18

250.000

108

200.000

24

350.000

114

150.000

30

540.000

120

100.000

36

735.000

126

72.000

42

1215.000

132

45.000

48

1800.000

138

25.000

54

1400.000

144

10.000

60

1050.000

150

0.000

66

900.000

156

0.000

72

740.000

162

0.000

78

620.000

   

84

510.000

   

 

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.

H (m)

O (m3/s)

S (m3)

130

20

1.00E+06

131

34

1.69E+06

132

57

2.85E+06

133

96

4.80E+06

134

162

8.12E+06

136

463

2.31E+07

137

781

3.91E+07

138

1318

6.59E+07

139

2226

1.11E+08

	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.

1

2

3

4

5

H (m)

O (m3/s)

S (m3)

2S/Dt (m3/s)

2S/Dt + O (m3/s)

130

20

1.00E+06

92.59259

112.5926

131

34

1.69E+06

156.4815

190.4815

132

57

2.85E+06

263.8889

320.8889

133

96

4.80E+06

444.4444

540.4444

134

162

8.12E+06

751.8519

913.8519

136

463

2.31E+07

2142.593

2605.593

137

781

3.91E+07

3615.741

4396.741

138

1318

6.59E+07

6103.704

7421.704

139

2226

1.11E+08

10303.7

12529.7

 

	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: So = 1'000,000 m3; Oo = 20 m3/s.

t = 6 -- i = 1

(Io + I1) = (0 + 50) m3/s = 50 m3/s

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

(2S1 /Dt + O1) = (Io + I1) + (2So /Dt - Oo) = 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