CE322 Basic Hydrology
Jorge A. Ramírez
Homework No. 8 - Solution
Problem 1. 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 |
90 |
450 |
|
6 |
50 |
96 |
330 |
|
12 |
120 |
102 |
280 |
|
18 |
225 |
108 |
210 |
|
24 |
300 |
114 |
160 |
|
30 |
525 |
120 |
110 |
|
36 |
700 |
126 |
90 |
|
42 |
1100 |
132 |
50 |
|
48 |
1500 |
138 |
30 |
|
54 |
1475 |
144 |
20 |
|
60 |
1300 |
150 |
15 |
|
66 |
1100 |
156 |
10 |
|
72 |
900 |
162 |
0 |
|
78 |
750 |
|
|
|
84 |
600 |
|
|
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 |
1000000 |
|
131 |
39 |
2150000 |
|
132 |
76 |
4300000 |
|
133 |
148 |
8450000 |
|
134 |
289 |
16600000 |
|
135 |
561 |
33000000 |
|
136 |
1092 |
64500000 |
|
137 |
2127 |
129000000 |
|
138 |
4143 |
257000000 |
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.
A. Develop the function 2S/Dt + O vs. O. Use a Dt of 1 hour, 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 |
1000000 |
92.59259 |
112.5926 |
|
131 |
39 |
2150000 |
199.0741 |
238.0741 |
|
132 |
76 |
4300000 |
398.1481 |
474.1482 |
|
133 |
148 |
8450000 |
782.4074 |
930.4074 |
|
134 |
289 |
1.66E+07 |
1537.037 |
1826.037 |
|
135 |
561 |
3.30E+07 |
3055.556 |
3616.556 |
|
136 |
1092 |
6.45E+07 |
5972.222 |
7064.222 |
|
137 |
2127 |
1.29E+08 |
11944.44 |
14071.44 |
|
138 |
4143 |
2.57E+08 |
23796.3 |
27939.3 |
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 O corresponding to the value of (2S1 /Dt + O1) obtained above. Use interpolation as indicated below.
O1 = 20 m3/s + [(39 - 20)/(238.07 - 112.5925)] (122.593 - 112.5925) m3/s = 21.514 m3/s
t = 12 - i = 2
(I1 + I2) = (50 + 120) m3/s = 170 m3/s
(2S1 /Dt - O1) = (2S1 /Dt + O1) - 2 x O1 = 122.593 m3/s - 2 x 21.514 m3/s = 79.564 m3/s
(2S2 /Dt + O2) = (I1 + I2) + (2S1 /Dt - O1) = 249.564m3/s
Using the relationship (2S/Dt + O) vs. O developed in Part A, obtain the outflow O corresponding to the value of (2S1 /Dt + O1) obtained above. Use interpolation as indicated below.
O1 = 39 m3/s + [(76 - 39)/(474.148 - 238.074)] (249.564 - 238.074) m3/s = 40.80 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) |
2Si+1/Dt - Oi+1 (m3/s) |
|
0 |
0 |
|
|
|
20 |
|
|
6 |
50 |
50 |
72.59259 |
122.5926 |
21.51417 |
79.56425 |
|
12 |
120 |
170 |
79.56425 |
249.5643 |
40.80086 |
167.9625 |
|
18 |
225 |
345 |
167.9625 |
512.9625 |
82.1251 |
348.7123 |
|
24 |
300 |
525 |
348.7123 |
873.7123 |
139.0532 |
595.6058 |
|
30 |
525 |
825 |
595.6058 |
1420.606 |
225.1725 |
970.2609 |
|
36 |
700 |
1225 |
970.2608 |
2195.261 |
345.0893 |
1505.082 |
|
42 |
1100 |
1800 |
1505.082 |
3305.082 |
513.6837 |
2277.715 |
|
48 |
1500 |
2600 |
2277.715 |
4877.715 |
755.2402 |
3367.234 |
|
54 |
1475 |
2975 |
3367.235 |
6342.234 |
980.8015 |
4380.631 |
|
60 |
1300 |
2775 |
4380.631 |
7155.631 |
1105.502 |
4944.628 |
|
66 |
1100 |
2400 |
4944.628 |
7344.628 |
1133.417 |
5077.794 |
|
72 |
900 |
2000 |
5077.794 |
7077.794 |
1094.005 |
4889.785 |
|
78 |
750 |
1650 |
4889.785 |
6539.785 |
1011.228 |
4517.33 |
|
84 |
600 |
1350 |
4517.33 |
5867.33 |
907.658 |
4052.014 |
|
90 |
450 |
1050 |
4052.014 |
5102.014 |
789.7861 |
3522.441 |
|
96 |
330 |
780 |
3522.442 |
4302.441 |
666.6382 |
2969.165 |
|
102 |
280 |
610 |
2969.165 |
3579.165 |
555.3199 |
2468.525 |
|
108 |
210 |
490 |
2468.525 |
2958.525 |
461.0377 |
2036.449 |
|
114 |
160 |
370 |
2036.449 |
2406.449 |
377.1712 |
1652.107 |
|
120 |
110 |
270 |
1652.107 |
1922.107 |
303.5941 |
1314.919 |
|
126 |
90 |
200 |
1314.919 |
1514.919 |
240.0203 |
1034.878 |
|
132 |
50 |
140 |
1034.878 |
1174.878 |
186.4873 |
801.9036 |
|
138 |
30 |
80 |
801.9036 |
881.9036 |
140.3459 |
601.2119 |
|
144 |
20 |
50 |
601.2119 |
651.2119 |
103.9415 |
443.3288 |
|
150 |
15 |
35 |
443.3287 |
478.3287 |
76.65971 |
325.0093 |
|
156 |
10 |
25 |
325.0093 |
350.0093 |
56.54366 |
236.922 |
|
162 |
0 |
10 |
236.922 |
246.922 |
40.38673 |
166.1485 |
|
168 |
0 |
0 |
166.1485 |
166.1485 |
28.10926 |
109.93 |
|
174 |
0 |
0 |
109.93 |
109.93 |
0 |
109.93 |
Problem 2. Using the information tabulated below for a river reach, estimate the Muskingum parameters k and x. The initial storage in the reach is 6'000,000 m3.
|
Time (d) |
Inflow (m3/s) |
Output (m3/s) |
|
|
|
|
|
1 |
180. |
160. |
|
2 |
270. |
200. |
|
3 |
420. |
280. |
|
4 |
650. |
415. |
|
5 |
890. |
590. |
|
6 |
1100. |
770. |
|
7 |
1270. |
950. |
|
8 |
1360. |
1090. |
|
9 |
1380. |
1180. |
|
10 |
1390. |
1250. |
|
11 |
1370. |
1280. |
|
12 |
1350. |
1290. |
|
13 |
1310. |
1300. |
|
14 |
1260. |
1280. |
|
15 |
1210. |
1250. |
|
16 |
1160. |
1220. |
|
17 |
1100. |
1190. |
|
18 |
1000. |
1150. |
|
19 |
950. |
1100. |
|
20 |
900. |
1040. |
|
21 |
790. |
980. |
|
22 |
710. |
920. |
|
23 |
650. |
860. |
|
24 |
590. |
790. |
|
25 |
510. |
710. |
|
26 |
450. |
650. |
|
27 |
380. |
590. |
|
28 |
300. |
510. |
The Muskingum routing procedure is used for systems that have Storage - Discharge relationships that are hysteretic. That is, for systems for which the outflow is not a unique function of storage. The S vs. O relationship for the river reach under consideration is graphed below.
A. Parameter Estimation
Graphical Procedure:
The graphical procedure consists in generating graphs of [xI + (1-x)O] vs. S for different values of x, arbitrarily selected such that 0 < x < 0.5. The optimal value of x is selected as that which produces the narrowest and straightest loop graph of [xI + (1-x)O] vs. S. The slope of the least squares linear fit to the resulting points is the estimate of k.
a) Generate accumulated storage in the system. Use continuity equation as follows:
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
x=0.25 |
x=0.35 |
x=0.1 |
x=0.17 |
|||||
|
Inflow,
I (m3/s) |
Outflow,O (m3/s) |
Ave.
Inflow (m3/s) |
Ave.
Outflow (m3/s) |
Storage (m3) |
Weighted
Average Flux xI +
(1-x)O (m3/s) |
|||
|
180 |
160 |
|
|
6000000 |
165 |
167 |
162 |
163.4027 |
|
270 |
200 |
225 |
180 |
9888000 |
217.5 |
224.5 |
207 |
211.9096 |
|
420 |
280 |
345 |
240 |
18960000 |
315 |
329 |
294 |
303.8192 |
|
650 |
415 |
535 |
347.5 |
35160000 |
473.75 |
497.25 |
438.5 |
454.9823 |
|
890 |
590 |
770 |
502.5 |
58272000 |
665 |
695 |
620 |
641.0412 |
|
1100 |
770 |
995 |
680 |
85488000 |
852.5 |
885.5 |
803 |
826.1453 |
|
1270 |
950 |
1185 |
860 |
1.14E+08 |
1030 |
1062 |
982 |
1004.444 |
|
1360 |
1090 |
1315 |
1020 |
1.39E+08 |
1157.5 |
1184.5 |
1117 |
1135.937 |
|
1380 |
1180 |
1370 |
1135 |
1.59E+08 |
1230 |
1250 |
1200 |
1214.027 |
|
1390 |
1250 |
1385 |
1215 |
1.74E+08 |
1285 |
1299 |
1264 |
1273.819 |
|
1370 |
1280 |
1380 |
1265 |
1.84E+08 |
1302.5 |
1311.5 |
1289 |
1295.312 |
|
1350 |
1290 |
1360 |
1285 |
1.9E+08 |
1305 |
1311 |
1296 |
1300.208 |
|
1310 |
1300 |
1330 |
1295 |
1.93E+08 |
1302.5 |
1303.5 |
1301 |
1301.701 |
|
1260 |
1280 |
1285 |
1290 |
1.93E+08 |
1275 |
1273 |
1278 |
1276.597 |
|
1210 |
1250 |
1235 |
1265 |
1.9E+08 |
1240 |
1236 |
1246 |
1243.194 |
|
1160 |
1220 |
1185 |
1235 |
1.86E+08 |
1205 |
1199 |
1214 |
1209.792 |
|
1100 |
1190 |
1130 |
1205 |
1.8E+08 |
1167.5 |
1158.5 |
1181 |
1174.688 |
|
1000 |
1150 |
1050 |
1170 |
1.69E+08 |
1112.5 |
1097.5 |
1135 |
1124.479 |
|
950 |
1100 |
975 |
1125 |
1.56E+08 |
1062.5 |
1047.5 |
1085 |
1074.479 |
|
900 |
1040 |
925 |
1070 |
1.44E+08 |
1005 |
991 |
1026 |
1016.181 |
|
790 |
980 |
845 |
1010 |
1.3E+08 |
932.5 |
913.5 |
961 |
947.6739 |
|
710 |
920 |
750 |
950 |
1.12E+08 |
867.5 |
846.5 |
899 |
884.2712 |
|
650 |
860 |
680 |
890 |
94128000 |
807.5 |
786.5 |
839 |
824.2712 |
|
590 |
790 |
620 |
825 |
76416000 |
740 |
720 |
770 |
755.9725 |
|
510 |
710 |
550 |
750 |
59136000 |
660 |
640 |
690 |
675.9725 |
|
450 |
650 |
480 |
680 |
41856000 |
600 |
580 |
630 |
615.9725 |
|
380 |
590 |
415 |
620 |
24144000 |
537.5 |
516.5 |
569 |
554.2712 |
|
300 |
510 |
340 |
550 |
6000000 |
457.5 |
436.5 |
489 |
474.2711 |
Columns 1 & 2 are given.
Columns 3 & 4 are the average inflow flux (Ii+1 + Ii)/2 and outflow flux (Oi+1 + Oi)/2, respectively.
Column 5 is the cumulative storage in the system obtained using the continuity equation below.
Columns 6 - 9 are the values of the weighted average flux [xI + (1-x)O] for different values of x. The graph of Columns 6 - 9 vs. Column 5 is shown below.
Based on these results, a value of x = 0.17 is selected. The best fit to the corresponding points yields a value of k = 2.13 h.
Least Squares Procedure
|
Inflow (m3/s) |
Outflow (m3/s) |
Storage |
O2 (m3/s)2 |
I2 (m3/s)2 |
OI (m3/s)2 |
SO (m6/s) |
SI (m6/s) |
|
180 |
160 |
6000000 |
25600 |
32400 |
28800 |
960000000 |
1080000000 |
|
270 |
200 |
9888000 |
40000 |
72900 |
54000 |
1977600000 |
2669760000 |
|
420 |
280 |
18960000 |
78400 |
176400 |
117600 |
5308800000 |
7963200000 |
|
650 |
415 |
35160000 |
172225 |
422500 |
269750 |
14591400000 |
22854000000 |
|
890 |
590 |
58272000 |
348100 |
792100 |
525100 |
34380480000 |
51862080000 |
|
1100 |
770 |
85488000 |
592900 |
1210000 |
847000 |
65825760000 |
94036800000 |
|
1270 |
950 |
113568000 |
902500 |
1612900 |
1206500 |
1.0789E+11 |
1.44231E+11 |
|
1360 |
1090 |
139056000 |
1188100 |
1849600 |
1482400 |
1.51571E+11 |
1.89116E+11 |
|
1380 |
1180 |
159360000 |
1392400 |
1904400 |
1628400 |
1.88045E+11 |
2.19917E+11 |
|
1390 |
1250 |
174048000 |
1562500 |
1932100 |
1737500 |
2.1756E+11 |
2.41927E+11 |
|
1370 |
1280 |
183984000 |
1638400 |
1876900 |
1753600 |
2.355E+11 |
2.52058E+11 |
|
1350 |
1290 |
190464000 |
1664100 |
1822500 |
1741500 |
2.45699E+11 |
2.57126E+11 |
|
1310 |
1300 |
193488000 |
1690000 |
1716100 |
1703000 |
2.51534E+11 |
2.53469E+11 |
|
1260 |
1280 |
193056000 |
1638400 |
1587600 |
1612800 |
2.47112E+11 |
2.43251E+11 |
|
1210 |
1250 |
190464000 |
1562500 |
1464100 |
1512500 |
2.3808E+11 |
2.30461E+11 |
|
1160 |
1220 |
186144000 |
1488400 |
1345600 |
1415200 |
2.27096E+11 |
2.15927E+11 |
|
1100 |
1190 |
179664000 |
1416100 |
1210000 |
1309000 |
2.138E+11 |
1.9763E+11 |
|
1000 |
1150 |
169296000 |
1322500 |
1000000 |
1150000 |
1.9469E+11 |
1.69296E+11 |
|
950 |
1100 |
156336000 |
1210000 |
902500 |
1045000 |
1.7197E+11 |
1.48519E+11 |
|
900 |
1040 |
143808000 |
1081600 |
810000 |
936000 |
1.4956E+11 |
1.29427E+11 |
|
790 |
980 |
129552000 |
960400 |
624100 |
774200 |
1.26961E+11 |
1.02346E+11 |
|
710 |
920 |
112272000 |
846400 |
504100 |
653200 |
1.0329E+11 |
79713120000 |
|
650 |
860 |
94128000 |
739600 |
422500 |
559000 |
80950080000 |
61183200000 |
|
590 |
790 |
76416000 |
624100 |
348100 |
466100 |
60368640000 |
45085440000 |
|
510 |
710 |
59136000 |
504100 |
260100 |
362100 |
41986560000 |
30159360000 |
|
450 |
650 |
41856000 |
422500 |
202500 |
292500 |
27206400000 |
18835200000 |
|
380 |
590 |
24144000 |
348100 |
144400 |
224200 |
14244960000 |
9174720000 |
|
300 |
510 |
6000000 |
260100 |
90000 |
153000 |
3060000000 |
1800000000 |
|
|
|
|
SO2= 25720025 |
SI2 = 26336400 |
SOI = 25559950 |
SSO = 3.42122E+12 |
SSI = 3.42112E+12 |
Using the above equations yields:
A = 22655.3838 s
B = 110503.2815 s
k = A+B = 133158.6653 s = 1.54 d
x = A/(A + B) = 0.17013
B. Muskingum Routing
Use the Muskingum routing procedure to route the original hydrograph.
Select a Dt = 1 d, as suggested by the inflow data. However, check that with the selected Dt, parameter values meet restrictions:
x < 0.5 Dt/k < 1 - x
For this case, we have 2 sets of parameters. However, both sets meet the parameter restrictions:
Graphical Procedure parameters:
0.17 < (0.5) (1 d)/2.13 d < 1 - 0.17 Thus, OK.
Least Squares Procedure parameters:
0.17 < (0.5) (1 d)/1.54 d < 1 - 0.17 Thus, OK. Proceed with routing, by obtaining Co, C1, and C2.
In what follows, the Least Squares parameters are used.
This yields: Co = 0.13366; C1 = 0.42846; and C2 = 0.437878. Using these values in the Muskingum routing equation:
obtain the outflow hydrograph as tabulated below.
|
Time (d) |
Inflow (m3/s) |
Co
x Ii+1 (m3/s) |
C1
x Ii (m3/s) |
C2
x Oi (m3/s) |
Outflow (m3/s) |
|
1 |
180 |
|
|
|
160 |
|
2 |
270 |
36.08955 |
77.12226 |
70.06048 |
183.2723 |
|
3 |
420 |
56.1393 |
115.6834 |
80.2509 |
252.0736 |
|
4 |
650 |
86.88225 |
179.9519 |
110.3775 |
377.2117 |
|
5 |
890 |
118.9619 |
278.4971 |
165.1727 |
562.6316 |
|
6 |
1100 |
147.0315 |
381.3267 |
246.364 |
774.7222 |
|
7 |
1270 |
169.7546 |
471.3027 |
339.2338 |
980.2911 |
|
8 |
1360 |
181.7844 |
544.1404 |
429.2479 |
1155.173 |
|
9 |
1380 |
184.4577 |
582.7015 |
505.8247 |
1272.984 |
|
10 |
1390 |
185.7944 |
591.2707 |
557.4117 |
1334.477 |
|
11 |
1370 |
183.1211 |
595.5552 |
584.338 |
1363.014 |
|
12 |
1350 |
180.4478 |
586.9861 |
596.834 |
1364.268 |
|
13 |
1310 |
175.1012 |
578.417 |
597.3829 |
1350.901 |
|
14 |
1260 |
168.4179 |
561.2787 |
591.5298 |
1321.226 |
|
15 |
1210 |
161.7347 |
539.8558 |
578.536 |
1280.126 |
|
16 |
1160 |
155.0514 |
518.433 |
560.5392 |
1234.024 |
|
17 |
1100 |
147.0315 |
497.0101 |
540.3518 |
1184.393 |
|
18 |
1000 |
133.665 |
471.3027 |
518.6198 |
1123.588 |
|
19 |
950 |
126.9818 |
428.457 |
491.9943 |
1047.433 |
|
20 |
900 |
120.2985 |
407.0342 |
458.6479 |
985.9805 |
|
21 |
790 |
105.5954 |
385.6113 |
431.7392 |
922.9458 |
|
22 |
710 |
94.90215 |
338.481 |
404.1377 |
837.5209 |
|
23 |
650 |
86.88225 |
304.2045 |
366.732 |
757.8187 |
|
24 |
590 |
78.86235 |
278.4971 |
331.8321 |
689.1915 |
|
25 |
510 |
68.16915 |
252.7896 |
301.7818 |
622.7406 |
|
26 |
450 |
60.14925 |
218.5131 |
272.6844 |
551.3467 |
|
27 |
380 |
50.7927 |
192.8057 |
241.4226 |
485.021 |
|
28 |
300 |
40.0995 |
162.8137 |
212.38 |
415.2932 |
The resulting hydrographs are graphed below.
