ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Problem 7.7.11
A triangular synthetic unit hydrograph developed by the Soil Conservation Service method has qp =
2,900 cfs/in, TP = 50 min, and tr = 10 min. Compute the direct runoff hydrograph for a 20-minute storm,
having 0.66 in rainfall in the first 10 minutes and 1.70 in. in the second 10 minutes. The rainfall loss rate
is Φ = 0.6 in/h throughout the storm.
10 Minute SCS Unit Hydrograph
3500
3000
cfs/in
2500
2000
1500
1000
500
0
0
0.5
1
1.5
2
2.5
Time, (hours)
Figure 10.1 10 Minute SCS Unit Hydrograph, Problem (7.7.11)
The SCS Unit Hydrograph was obtained with the given data. Excess rainfall runoff was determined by
subtracting the rainfall loss rate per rainfall duration. The unit hydrograph ordinates were taken from
Figure 10.1 provided above. The direct runoff for the 20-minute storm was calculated per equation
(7.4.1) in the text and shown in Table 10.1 below:
1
ENV 6932 Advanced Env Hydrology
Time
(X10-min)
n=9
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Total:
Excess
Precipitation
(in)
0.56
1.60
Homework # 10 11/04/09
Unit hydrograph ordinates (cfs/in)
10 20
30
40
50
60
580 1160 1740 2320 2900 2552
325
928 650
1856
974
2784 1299
3712 1624
4640 1429
4083
2.2
70
2205
1235
3528
80
1858
1040
2973
Ryan Locicero #50779799
90
1510
100
1163
110
816
120
468
130
121
845.6
2416 651.28
1860.8 456.96
1305.6 262.08
748.8 67.76
Direct
runoff
(cfs)
325
1578
2830
4083
5336
6069
5318
4568
3818
3067
2318
1568
817
Total
: 41,695
Table 10.1 Direct Runoff Calculations (Problem 7.7.11)
2
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Problem 8.2.3
Solve for problem 8.2.2 assuming the initial reservoir storage is 87.5 X 106 m3.
Use the level pool routing method to route the hydrograph below through the reservoir whose storageoutflow characteristics are given in Prob. 8.2.1. What is the maximum reservoir discharge and storage?
Design Hydrograph Data
Time (h)
0
2
4
6
8
10
12
14
16
18
Inflow (m3/sec)
60
100
232
300
520
1,310
1,930
1,460
930
650
Reservoir Storage Characteristics:
Storage (106 m3)
75
81
87.5
100
110.2
Outflow (m3)
57
227
519
1330
2270
Δt = 2 hr or 7200 s.
This problem is solved by first determining the Storage-outflow function, this function is described
below:
2𝑆
+ 𝑄
𝛥𝑡
Example illustrating Storage-outflow function with storage, S = 87.5 x 106 m3 and Q = 519 m3/s
2 ∗ 87.5 x 106 m3
m3
+ 519
= 24,824 m3/s
7,200 𝑠
s
See Table 10.2 for a complete list of Storage-outflow function calculations
Table 10.2 Storage-outflow function, Problem (8.2.3)
Discharge
Q
Storage
S
(2S/Δt) +Q
(m3/s)
0
57
227
519
1330
2270
(X106 m3)
0
75
81
87.5
100
110.2
m3/s
20,890.33
22,727.00
24,824.56
29,107.78
32,881.11
3
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
In this problem the initial reservoir storage was 87.5 X 106 m3 resulting in an initial discharge of 519
m3/2. The corresponding storage-outflow function value provided an initial
2𝑆𝑗+1
𝛥𝑡
+ 𝑄𝑗+1 as shown in
Table 10.3. With the outflow value then known we can calculate the next unknown variable and use the
Storage-outflow function to determine the subsequent outflow value. These steps are outlined below.
2𝑆2
2𝑆2
− 𝑄2 = (
+ 𝑄2 ) − 2𝑄2
𝛥𝑡
𝛥𝑡
2𝑆2
− 𝑄2 = (24,984 − 2 ∗ 549) = 23,886 𝑐𝑚𝑠
𝛥𝑡
This is the value in column 5. This value is then added to the value in column 4 (j+1), to determine the
value for column 6 (j+1). The value generated in column 6 is then referenced to the storage outflow
function to determine an Outflow value, column 7. This is repeated until all inflow data is analyzed. See
Table 10.3 below:
Table 10.3 Routing of a flow through a detention reservoir by the level pool method. Problem (8.2.3)
Column:
1
2
3
4
5
6
7
Time
Time
Inflow
Ij + Ij+1
(2Sj/Δt)- Q
(2Sj+1/Δt) + Qj+1
Outflow
index j
(hr)
(cms)
(cms)
(cms)
(cms)
(cms)
1
0
60
60
24,824.00
2
2
100
160
23,886.00
24,984
549
3
4
232
332
23,350.00
24,218
434
4
6
300
532
23,108.00
23,882
387
5
8
520
820
23,140.00
23,928
394
6
10
1,310
1,830
23,878.00
24,970
546
7
12
1,930
3,240
25,212.00
27,118
953
8
14
1,460
3,390
26,134.00
28,602
1,234
9
16
930
2,390
26,086.00
28,524
1,219
10
18
650
1,580
25,552.00
27,666
1,057
519
The maximum Outflow from the data is given in Table 10.3 as 1,234 m3/s. This relates to a total storage
of 98.5 X 106 m3 determined interpolating the reservoir storage characteristics given.
4
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Problem 8.2.4
Solve Example 8.2.1 in the text if the initial depth in the reservoir is 2 ft. How much higher does this
make the maximum water level in the reservoir compared with the level found in Example 8.2.1?
Problem 8.2.4 was solved with the same procedure as the previous problem (8.2.3) with the data given
in Example 8.2.1. The initial reservoir depth of 2 ft was used resulting in a t=0 outflow rate of 30 cfs and
column 5 equal to 320 cfs as calculated in Table 10.4. See Tables below for calculated data.
Table 10.4 Storage-outflow function for determined detention reservoir (Problem 8.2.4)
1
Elevation
H
ft
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
2
Discharge
Q
cfs
3
8
17
30
43
60
78
97
117
137
156
173
190
205
218
231
242
253
264
275
3
Storage
S
ft3
21,780
43,560
65,340
87,120
108,900
130,680
152,460
174,240
196,020
217,800
239,580
261,360
283,140
304,920
326,700
348,480
370,260
392,040
413,820
435,600
4
(2S/Δt) +Q
cfs
76
153
235
320
406
496
586
678
770
863
955
1,044
1,134
1,221
1,307
1,393
1,476
1,560
1,643
1,727
5
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Table 10.5 Routing of flow through a detention reservoir by the level pool method (Problem 8.2.4)
Column:
1
2
Time
3
4
Time
Inflow
(min)
0
(cfs)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
10
20
30
40
50
60
70
80
90
100
110
120
130
140
16
17
18
19
20
21
22
150
160
170
180
190
200
210
index j
5
Ij + Ij+1
(cfs)
60
-
(2Sj/Δt)- Q
7
(2Sj+1/Δt) + Qj+1
Outflow
-
(cfs)
320.00
180
300
420
540
660
680
600
520
440
360
280
200
120
302.00
367.20
479.00
610.00
764.40
954.20
1,108.80
1,163.80
1,145.40
1,072.80
960.40
824.80
686.40
556.80
380.0
482.0
667.2
899.0
1,150.0
1,424.4
1,634.2
1,708.8
1,683.8
1,585.4
1,432.8
1,240.4
1,024.8
806.4
39.0
57.4
94.1
144.5
192.8
235.1
262.7
272.5
269.2
256.3
236.2
207.8
169.2
124.8
-
436.40
339.00
273.36
227.76
196.16
170.76
151.06
596.8
436.4
339.0
273.4
227.8
196.2
170.8
80.2
48.7
32.8
22.8
15.8
12.7
9.9
60
120
180
240
300
360
320
280
240
200
160
120
80
40
6
40
(cfs)
(cfs)
30.0
The maximum water level is calculated with by interpolating the values calculated in Table 10.4 at the
maximum outflow value of 272.5 cfs resulting in a maximum stage or water level of 9.89 ft. an increase
of 0.12 ft or 1.44 inches. The max water level calculated in the example is 9.77 ft.
6
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Problem 8.2.6
Consider a 2-acre detention basin with vertical walls. The triangular inflow hydrograph increases linearly
from zero to a peak of 540 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 8.2.2. The pipe is located at the bottom of the basin. Assuming the basin is
initially empty, use the level pool routing procedure with a 10-minute time interval to determine the
maximum depth in the detention basin.
The routing is set up using the specifications given in the problem. The storage outflow function data
for a 5 ft RCP was taken from Table 8.2.2. A Discharge-Elevation Relationship was developed for values
that were outside of the given table. A trendline was set to this relationship and provided an equation
for which additional storage outflow relationship was analyzed.
Discharge Elevation Relationship
300
y = 0.0439x4 - 1.2741x3 + 11.46x2 - 3.672x
Discharge, Q (cfs)
250
200
150
100
50
0
0.0
2.0
4.0
6.0
8.0
10.0
12.0
Elevation, H (ft)
Figure 10.2 Dishcarge-Elevation Relationship for 5ft Diameter RCP
Problem 8.2.6 was solved with the same procedure as the previous problem (8.2.3). See Table 10.6 and
10.7 for the Storage-outflow function for a given reservoir and routing of flow through a detention
reservoir by the level pool method.
7
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Table 10.6 Storage-outflow function (problem 8.2.6)
1
Elevation
H
2
Discharge
Q
3
Storage
S
4
(2S/Δt) +Q
ft
cfs
ft3
cfs
0.5
1.0
1.5
2.0
2.5
3.0
3
8
17
30
43
60
43,560
87,120
130,680
174,240
217,800
261,360
148
298
453
611
769
931
3.5
78
304,920
1,094
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
97
117
137
156
173
190
205
218
231
242
253
264
275
284
293
304
315
328
344
363
385
413
348,480
392,040
435,600
479,160
522,720
566,280
609,840
653,400
696,960
740,520
784,080
827,640
871,200
914,760
958,320
1,001,880
1,045,440
1,089,000
1,132,560
1,176,120
1,219,680
1,263,240
1,259
1,424
1,589
1,753
1,915
2,078
2,238
2,396
2,554
2,710
2,867
3,023
3,179
3,333
3,488
3,643
3,800
3,958
4,119
4,283
4,451
4,624
8
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Table 10.7 Routing of flow through a detention reservoir by the level pool method, Problem (8.2.6)
Column:
1
2
Time
3
4
5
6
7
Time
Inflow
Ij + Ij+1
(2Sj/Δt)- Q
(2Sj+1/Δt) + Qj+1
Outflow
(cfs)
(cfs)
(cfs)
(cfs)
(cfs)
1
(min)
0
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
19
20
21
22
180
190
200
210
index j
90
90
180
270
360
450
540
495
450
405
360
315
270
225
180
135
90
45
-
270
450
630
810
990
1,035
945
855
765
675
585
495
405
315
225
135
45
-
-
-
86.36
333.56
694.56
1,114.76
1,576.96
2,103.16
2,593.94
2,945.74
3,170.74
3,283.14
3,302.14
3,242.54
3,116.74
2,930.74
2,688.34
2,400.74
2,076.94
90.0
356.4
783.6
1,324.6
1,924.8
2,567.0
3,138.2
3,538.9
3,800.7
3,935.7
3,958.1
3,887.1
3,737.5
3,521.7
3,245.7
2,913.3
2,535.7
1.8
11.4
44.5
104.9
173.9
231.9
272.1
296.6
315.0
326.3
328.0
322.3
310.4
295.5
278.7
256.3
229.4
1,733.94
1,427.94
1,193.14
1,014.32
2,121.9
1,733.9
1,427.9
1,193.1
194.0
153.0
117.4
89.4
The maximum depth was determined by interpolating the storage-outflow function relationship relating
to Elevation. The maximum outflow of 328 cfs results in a maximum depth of 12.5 ft
9
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Problem 8.3.8
In this problem, you are to determine the runoff from a particular watershed and route the runoff
hydrograph through a reservoir at the downstream end of the watershed. The reservoir has the
following storage-outflow characteristics:
Storage (ac*ft)
Outflow (cfs)
0
0
200
2
300
20
400
200
500
300
600
350
700
450
1100
1200
0.0
0.5
1.0
1.5
2.0
0.0
1.0
3.0
4.0
4.5
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
0
4.00
200
4.50
500
5.00
800
5.50
700
6.00
600
500
400
300
200
100
50
0
The rainfall is:
Time (h)
Accumulated rainfall depth
(in)
The half hour unit hydrograph is:
Time (h)
Discharge
(cfs/in)
Time (h)
Discharge
(cfs/in)
The phi-index is 0.8 in/hr is to be used to account for losses. Determine the peak discharge from the
reservoir assuming zero baseflow. What is the area in square miles of the watershed?
The unit hydrograph data was used to determine the direct runoff for the watershed. To account for
abstractions the Phi-Index was subtracted from accumulated rainfall depth data before plugging it into
the spreadsheet for calculation. See Table 10.8
10
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Table 10.8 Calculation for Direct Runoff Hydrograph, (Problem 8.3.8)
Time (0.5
hours) n=9
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Total:
Excess
Precipitation 0.5
(in)
200
0.60
120
2.60
520
3.60
720
4.10
820
Unit hydrograph ordinates (cfs/in)
1.0
1.5
2.0
2.5
500
800
700
600
300
1300
1800
2050
480
2080
2880
3280
420
1820
2520
2870
360
1560
2160
2460
3.0
500
300
1300
1800
2050
3.5
400
240
1040
1440
1640
4.0
300
180
780
1080
1230
4.5
200
120
520
720
820
10.9
5.0
100
60
260
360
410
5.5
50
30
130
180
205
Direct
runoff
(cfs)
120
820
2500
5120
7110
7660
6570
5480
4390
3300
2240
1310
590
205
Total : 47,415
The storage-outflow function was determined and extrapolated for use in this problem. See Table 10.9 and Figure 10.3
11
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Discharge-Storage Relationship
1,400
1,200
Discharge, Q (cfs)
1,000
800
600
400
200
-
200
400
(200)
600
800
1,000
1,200
Storage (ac-ft)
Figure 10.3 Discharge-Storage Relationship, Problem 8.3.8
Table 10.9 Storage-outflow function, Problem 8.3.8
1
Discharge
Q
cfs
2
Storage
S
(ac-ft)
-
3
(2S/Δt) +Q
cfs
2
20
200
300
350
450
200
300
400
500
600
700
9,682
14,540
19,560
24,500
29,390
34,330
1200
1,100
54,440
1950
1,500
74,550
Problem 8.3.8 was solved with the same procedure as the previous problem and provided examples as
illustrated in Problem (8.2.3).
12
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
Ryan Locicero #50779799
Table 10.10 Routing of a flow through a detention reservoir by the level pool method, Problem 8.3.8
Column:
1
2
Time
3
4
5
6
7
Time
Inflow
Ij + Ij+1
(2Sj/Δt)- Q
(2Sj+1/Δt) + Qj+1
Outflow
(cfs)
(cfs)
(cfs)
(cfs)
(cfs)
1
2
3
(hr)
0.0
0.5
1.0
4
1.5
5
6
7
8
9
2.0
2.5
3.0
3.5
4.0
10
4.5
11
5.0
12
5.5
13
6.0
14
15
16
17
18
19
20
21
22
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
index j
120
820
2,500
5,120
7,110
7,660
6,570
5,480
4,390
3,300
2,240
1,310
590
120
940
119.96
1,059.52
120.0
1,060.0
0.0
0.2
4,377.72
4,379.5
0.9
7,620
12,230
14,770
14,230
12,050
11,976.56
23,612.56
37,180.56
49,236.56
58,376.56
11,997.7
24,206.6
38,382.6
51,410.6
61,286.6
10.6
297.0
601.0
1,087.0
1,455.0
9,870
64,818.56
68,246.6
1,714.0
7,690
68,762.56
72,508.6
1,873.0
5,540
70,422.56
74,302.6
1,940
3,550
70,116.56
73,972.6
1,928.0
68,306.56
65,609.56
62,566.56
59,560.56
56,780.56
54,206.56
51,824.56
49,620.56
47,580.56
72,016.6
69,101.6
65,814.6
62,566.6
59,560.6
56,780.6
54,206.6
51,824.6
49,620.6
1,855.0
1,746.0
1,624.0
1,503.0
1,390.0
1,287.0
1,191.0
1,102.0
1,020.0
3,320
1,900
205
-
795
205
-
The peak discharge from the reservoir was calculated as 1,940 cfs as shown in Table 10.10.
What is the area of the watershed in problem 8.3.8?
The summation of the direct runoff = 47,415 cfs (Table 10.8)
𝑁
𝑉𝑑 = ∑ 𝑄𝑛 𝛥𝑡
𝑛=1
13
ENV 6932 Advanced Env Hydrology
Homework # 10 11/04/09
𝑉𝑑 = 47,415
Ryan Locicero #50779799
𝑓𝑡 3
𝑠
∗ 1 ℎ𝑟 ∗ 3600
𝑠
ℎ𝑟
𝑉𝑑 = 1.71 𝑋 108 𝑓𝑡 3
Depth of direct runoff, rd = 10.9 in = 0.91 ft (Table 10.8)
𝐴=
𝐴=
𝑉𝑑
𝑟𝑑
1.71 𝑋 108 𝑓𝑡 3
. 91 𝑓𝑡
𝐴 = 1.87 𝑋 108 𝑓𝑡 2
𝐴=
1.87 𝑋 108 𝑓𝑡 2
5,280 2 𝑓𝑡 2 /𝑚𝑖 2
𝑨 = 𝟔. 𝟕𝟒 𝒎𝒊𝟐
14