# Solution Manual for Hydrology and Floodplain Analysis, 6th Edition

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Chapter 2-ANSWER KEY
Concept Check
2.1 What characterizes a watershed?
ANSWER:
A watershed is the basic unit used in most hydrologic calculations relating to the
water balance or computation of rainfall-runoff. It is characterized by one main
channel and by tributaries that drain into a main channel at one or more confluence
points. Larger watersheds can have many subareas that contribute runoff to a single
outlet. The drainage area is another characteristic that reflects the volume of water
that can be generated from rainfall.
2-1
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2.2 Describe the four assumptions made that are inherent to the unit hydrograph.
ANSWER:
1. Rainfall excesses of equal duration are assumed to produce hydrographs with
equivalent time bases regardless of the intensity of the rain.
2. Direct runoff ordinates for a storm if given duration are assumed directly
proportional to rainfall excess volumes. Thus, twice the rainfall produces a
doubling hydrograph ordinates.
3. The time distribution of direct runoff is assumed independent of antecedent
precipitation.
4.Rainfall distribution is assumed to be the same for all storms of equl duration, both
spatially and temporally.
These assumptions can sometimes limit the application of a unit hydrograph in a
given watershed.
2-2
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2.3 What physical factors affect the shape and timing of the unit hydrograph?
ANSWER:
The shape and timing of the hydrograph are related to duration and intensity of
rainfall as well as the various factors governing the watershed area.
The meteorological factors that influence the hydrograph shape and volume of runoff
include:
1.Rainfall intensity and pattern
2.Areal distribution of rainfall over the basin
The various factors that govern the watershed include:
1.Size and shape of the drainage area
2.Slope of the land surface and the main channel
3.Channel morphology and drainage type
4.Soil types and distribution
5.Storage detention in the watershed.
2-3
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2.4 Discuss the parameters that describe the main timing aspects of the hydrograph.
ANSWER:
The physical factors that will affect the shape and timing of the unit hydrograph are:
a. Duration of the rainfall excess: the time from start to finish of rainfall excess
b. Lag Time: the time from the center of mass of rainfall excess to the peak of the
hydrograph.
c. Time of rise: the time from the start of rainfall excess to the peak of the
hydrograph.
d. Time of concentration: the time for a wave (of water) to propagate from the
most distant point in the watershed to the outlet. One estimate is the time from
the end of net rainfall to the inflection point of the hydrograph.
2-4
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2.5 What two characteristics of the hydrograph do most methods for synthetic unit
hydrographs relate?
ANSWER:
They relate hydrograph peak flow and timing to watershed characteristics.
2-5
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2.6
In order to determine which of the three excavation sites could have potentially
contributed to the sedimentation of the three lakes, it is necessary to determine the
watershed boundaries and the contributing drainage areas of the system of lakes.
Catchment basins (watersheds) can be determined by connecting ridge lines and
dividing lines. Ridge lines will follow the elevation isocontours and dividing lines
are perpendicular to the elevation isocontours at the end and start at the confluences
of streams.Using the map on the textbook website displaying the elevation data,
determine the following:
a. How many confluences are displayed on the map?
b. Delineate the watersheds using the elevation isocontours. The total drainage area
of the study area is approximately 238 acres.
c. Determine which of the three lakes may be affected by accelerated sedimentation.
d. Determine which of the three excavation sites may be responsible for the
accelerated sedimentation of the lakes.
ANSWER:
a. There are three confluence points
b. See picture below
c. Lakes 1 (lake 2 minimally) will be affected by the accelerated excavation.
d. The sites that are responsible are sites 1 and 2. Three can be eliminated because it
is outside the watershed.
2-6
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2.6 contโ
HOMEWORK PROBLEMS
2.7
A watershed has the following characteristics:
2-7
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A = 2500 ac
L = 3.9 mi.
S = 52 ft/mi.
y = 1%
The channel is lined with concrete.
The watershed is a residential area with 1/4-ac lots. The soil is categorized as soil
group B. Assume that the average watershed slope is the same as the channel slope.
Determine the UH for this area for a storm duration of 1 hr using the SCS triangular
UH method.
ANSWER:
From table 2-1 obtain the curve number with the given information.
CN = 75
S = (1000/CN) โ 10 = (1000/75) โ 10 = 3.33
L = 3.9 mi. = (3.9mi.)(5280 ft/mi.) = 20592 ft.
Plug into equation:
tp ๏ฝ
tp ๏ฝ
L0.8 ( S ๏ซ 1) 0.7
1900 y
(21,120) 0.8 (3.33 ๏ซ 1) 0.7
1900 1
= 4.15 hr
TR = D/2 + tp = 1/2 + 4.15 = 4.65 hr
Qp ๏ฝ
484 A
๏ฝ
TR
484 ๏ 2600ac ๏
4.73
1mi.2
640ac ๏ฝ 406 cfs
2.7 contโ
Vol = (1 in.)(2500 ac) = 2500 ac-in. ~ 2500 cfs-hr
2-8
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Vol = (1/2)(Qp)(TR + B)
2Vol
๏ญ TR = 7.67 hr
QP
Flow (cfs)
B๏ฝ
450
0
400
0.5
350
1
1.5
300
2
250
2.5
200
3
150
3.5
100
4
50
4.5
0
5
0
2
4
6
8
10
12
14
Time (hr)
2-9
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2.8
Determine the storm hydrograph resulting from the rainfall pattern in Fig. P2โ8(a)
using the triangular 1-hr UH given in Fig. P2โ8(b).
Change Rainfall to 1, 2, 3, 2
Fig. P2-8a
Fig. P2-8b
ANSWER:
Since the rainfall is given in net rainfall, we do not need to subtract any losses. The
rainfall for each time period is multiplied by the unit hydrograph ordinates to obtain the
storm hydrograph. This computational procedure is in the table below:
2.8 contโ
2-10
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In order to obtain R 1 U through R 4 U. You take the UH units and you multiply it by the
intensity for that given hour. Once you proceed to the next hour the values are lagged one
hour. To obtain Q the values are then added across the table.
Time
(hrs)
0
1
2
3
4
5
6
7
8
9
R2 U
R1 U
UH
0
0.5
1
0.75
0.5
0.25
0
0
0.5
1
0.75
0.5
0.25
0
1
2
1.5
1
0.5
0
R3U
R4 U
0
1.5
3
2.25
1.5
0.75
0
Q
0
0.5
2
4.25
6
5.5
3.5
1.75
0.5
0
0
1
2
1.5
1
0.5
0
The hydrograph can then be plotted (below is new hydrograph)
7
6
Flow (cfs)
5
4
3
2
1
0
0
2
4
6
8
10
Time (hr)
2-11
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2.9
(a) Given a triangular 1-hr UH with
TB ๏ฝ 12 hr,
TR ๏ฝ 4 hr,
QP ๏ฝ 200 cfs,
where
TB ๏ฝ time base of the UH,
TR ๏ฝ time of rise,
QP ๏ฝ peak flow,
develop a storm hydrograph for hourly rainfall (in.) of P ๏ฝ ๏ฉ๏ซ0.1, 0.5, 1.2 ๏น๏ป .
(b)
Repeat the above problem for hourly rainfall (in.) of P ๏ฝ ๏ฉ๏ซ0.2, 1.0, 2.4 ๏น๏ป .
ANSWER:
The UH ordinates are derived using the triangular 1-hr UH graph in 1 hour intervals:
2.9 contโ
2-12
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Time
(hr) U (cfs)
0
0
1
50
2
100
3
150
4
200
5
175
6
150
7
125
8
100
9
75
10
50
11
25
12
0
The storm hydrograph is developed for both cases (a) and (b) with the use of the
convolution equation.
2.9 contโ
2-13
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a.
Time
(hr)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
U (cfs)
0
50
100
150
200
175
150
125
100
75
50
25
0
P1*U
0
5
10
15
20
17.5
15
12.5
10
7.5
5
2.5
0
P2*U
0
25
50
75
100
87.5
75
62.5
50
37.5
25
12.5
0
P3*U
0
60
120
180
240
210
180
150
120
90
60
30
0
Q (cfs)
0
5
35
125
215
297.5
342.5
297.5
252.5
207.5
162.5
117.5
72.5
30
0
2-14
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2.9 contโ
b.
Time
(hr)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
U
(cfs)
0
50
100
150
200
175
150
125
100
75
50
25
0
P1*U
P2*U
0
10
20
30
40
35
30
25
20
15
10
5
0
0
50
100
150
200
175
150
125
100
75
50
25
0
P3*U
0
120
240
360
480
420
360
300
240
180
120
60
0
Q
0
10
70
250
430
595
685
595
505
415
325
235
145
60
0
2-15
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2.9 contโ
2-16
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2.10 A sketch of the Buffalo Creek Watershed is shown in Fig. P2โ10. Areas A and B
are identical in size, shape, slope, and channel length. UHs (1 hr) are provided for
natural and fully developed conditions for both areas.
Fig P2-10
(a) Assuming natural conditions for both areas, evaluate the peak outflow at point 1
if 2.5 in./hr of rain falls for 2 hr. Assume a total infiltration loss of 1 in.
(b) Assume that area B has reached full development and area A has remained in
natural conditions. Determine the outflow hydrograph at point 1 if a net rainfall
of 2 in./hr falls for 1 hr.
(c) Sketch the outflow hydrograph for the Buffalo Creek Watershed under
complete development (A and B both urbanized) for the rainfall given in part
(b).
Time (hr)
0
1
2
3
4
5
6
7
8
UHdev (cfs)
0
40
196
290
268
185
90
30
0
Time (hr)
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14
UH (nat)
0
12
32
62
108
180
208
182
126
80
53 32 18 6
2.10 contโ
2-17
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0
ANSWER:
a. Assuming uniform loss, there is a loss rate of:
1 in. for two hours =
Net rainfall intensity = 2.5 โ 0.5 = 2 in/hr for two hours
The storm hydrograph at point 1 is found by combining the storm hydrographs for
areas A and B.
Existing conditions for both areas:
Time
(hr)
(cfs)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
P1*U
0
12
32
62
108
180
208
182
126
80
53
32
18
6
0
0
0
24
64
124
216
360
416
364
252
160
106
64
36
12
0
0
P2*U
0
24
64
124
216
360
416
364
252
160
106
64
36
12
0
(cfs)
0
24
88
188
340
576
776
780
616
412
266
170
100
48
12
0
(cfs) Q (cfs)
0
0
24
48
88
176
188
376
340
680
576
1152
776
1552
780
1560
616
1232
412
824
266
532
170
340
100
200
48
96
12
24
0
0
2-18
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2.10 contโ
Storm hydrograph at point 1: Existing
conditions
1800
1600
Flow (cfs)
1400
1200
1000
800
600
400
200
0
0
2
4
6
8
10
12
14
16
Time (hr)
b. In this case we have to determine the storm hydrograph at point 1 for a 1-hr
duration rainfall. i = 2in/hr for 1 hr
2.10 contโ
2-19
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Time (hr)
Unat (cfs)
Udevel
Qa (cfs)
Qb (cfs)
Q (cfs)
0
0
0
0
0
0
1
12
40
24
80
104
2
32
196
64
392
456
3
62
290
124
580
704
4
108
268
216
536
752
5
180
185
360
370
730
6
208
90
416
180
596
7
182
30
364
60
424
8
126
0
252
0
252
9
80
160
160
10
53
106
106
11
32
64
64
12
18
36
36
13
6
12
12
14
0
0
0
15
0
0
0
2.10 contโ
2-20
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c. Areas A and B are fully developed. i = 2in/hr for 1 hr
Time
(hr)
Udevel
Qa (cfs)
Qb (cfs)
Q (cfs)
0
0
0
0
0
1
40
80
80
160
2
196
392
392
784
3
290
580
580
1160
4
268
536
536
1072
5
185
370
370
740
6
90
180
180
360
7
30
60
60
120
8
0
0
0
0
PROBLEM 2.11
2-21
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(a) Examine the two hydrographs presented below for an area of 5000 acres. Which of the two is
a unit hydrograph? Please show how you arrived at this conclusion.
(b) Using the hydrograph that you chose from part (a), assuming it is a 1-hr UH, create a 3-hr UH.
1400
1200
Q (cfs
1000
800
Hydrograph 1
600
Hydrograph 2
400
200
0
0
2
4
6
8
10
12
Time (hr)
2-22
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SOLUTION 2.11
(a)
In order to determine this, you need to add the ordinates of each of the individual hydrographs
Time (hr)
Hydrograph 1
Hydrograph 2
1
0
0
2
75
200
3
250
500
4
450
700
5
600
1000
6
500
1300
7
350
700
8
225
400
9
50
200
10
0
0
2500
5000
Then multiply this cfs value by the time step, in this case 1 hour meaning that hydrograph
1 has a runoff volume of 2500 cfs-hrs and hydrograph 2 has a runoff volume of 5000 cfs-hr.
Assuming that the conversion between cfs-hrs and acre-inches is a 1:1 relationship we now have
the volumes in ac-in. To get the inches of direct runoff we divide these values by the size of the
watershed, in this case 5000 acres to get the inches of direct runoff. Because hydrograph 2
corresponds to 1 inch of direct runoff while hydrograph 1 corresponds to ยฝ inch of direct runoff,
hydrograph 2 is the unit hydrograph.
(b) Using hydrograph two as our 1-hr UH we can develop the 3-hr UH.
2.11 contโ
2-23
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Time (hr)
1-hr UH
Lagged 1-hr
UH
0
0
1
200
0
2
500
200
3
700
4
Sum
3-hr UH
0
0
200
66.67
0
700
233.33
500
200
1400
466.67
1000
700
500
2200
733.33
5
1300
1000
700
3000
1000
6
700
1300
1000
3000
1000
7
400
700
1300
2400
800
8
200
400
700
1300
433.33
9
0
200
400
600
200
0
200
200
66.67
0
0
0
10
Lagged 1-hr UH
11
3-hr UH
1200
Q (cfs)
1000
800
600
400
200
0
0
2
4
6
8
10
12
Time (hr)
2-24
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2.12 A major storm event was recorded for Little Bear Creek. The incremental rainfall
and measured hydrograph data for this storm are provided in Table P2โ12 in 1-hr
increments. The drainage area for the basin is 3.25 mi2. Assume base flow for Little
Bear is zero. (See Fig. 1.26).
(a) Using the storm hydrograph, estimate the volume of runoff that occurred in
inches over the watershed.
(b) Estimate the volume of infiltration in inches for this storm event based on the
measured rainfall.
(c) Estimate the time to peak t p for this watershed for the entire storm event.
Table P2โ12. Little Bear Creek Data
Date/Time
Incremental
Precipitation (in.)
Flow (cfs)
6/8/01
16:00
0
0.0
6/8/01
17:00
0.05
0.5
6/8/01
18:00
0.04
0.5
6/8/01
19:00
0
0.6
6/8/01
20:00
0.08
14
6/8/01
21:00
0.05
32
6/8/01
22:00
0.42
54
6/8/01
23:00
0.1
78
6/9/01
0:00
0.3
105
6/9/01
1:00
0.14
158
6/9/01
2:00
0.19
201
2-25
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6/9/01
3:00
0.45
228
6/9/01
4:00
0.01
258
6/9/01
5:00
0
272
6/9/01
6:00
225
6/9/01
7:00
189
6/9/01
8:00
142
6/9/01
9:00
119
6/9/01
10:00
98
6/9/01
11:00
66
6/9/01
12:00
42
6/9/01
13:00
27
6/9/01
14:00
0
ANSWER:
a. The volume can be determined be using the individual flows, timing it by the time
interval and then adding them. The volume will be in cfs-hr. In this case since the
time interval is 1 hr.
2309.6 cfs-hr
2309.6 ac-in
We are given the area in square miles which can be converted to acres and can be
used to find runoff in inches.
3.25 mi
2
b. Knowing the amount of direct runoff in inches we can then subtract it from total
rainfall to find what was lost to infiltration:
With the incremental precipitation we can find the total rainfall =
.
Thus, total precipitation โ direct runoff = infiltration loss
2.12 contโ
2-26
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c. Time to peak is found knowing the center of mass of rainfall excess and the time
to the peak of the hydrograph. The peak of the hydrograph happens 13 hrs after
the storm starts (at 5 oโclock). The center of mass of rainfall can be calculated by
finding the weighted average of the rain (multiplying rainfall by the time) and
then dividing it by the total rainfall.
See Excel table below.
13
13
i ๏ฝ0
i ๏ฝ0
๏ฅ Pi t i / ๏ฅ Pi ๏ฝ Time to Center of Mass of Rainfall
14.55in. ๏ hr
= 7.95 hr
1.83in.
Time to Peak Flow โ Time to CM of Rainfall = 13 โ 7.95 = 5 hr = Time to Peak
Note: โTime to Peakโ is NOT equal to โTime to Peak Flowโ
Date/Time
Time Increment, t (hr)
Precipitation, P (in.)
P*t
Flow (cfs)
6/8/2001
16:00
0
0
0
0
6/8/2001
17:00
1
0.05
0.05
0.5
6/8/2001
18:00
2
0.04
0.08
0.5
6/8/2001
19:00
3
0
0
0.6
6/8/2001
20:00
4
0.08
0.32
14
6/8/2001
21:00
5
0.05
0.25
32
6/8/2001
22:00
6
0.42
2.52
54
6/8/2001
23:00
7
0.1
0.7
78
6/9/2001
0:00
8
0.3
2.4
105
6/9/2001
1:00
9
0.14
1.26
158
6/9/2001
2:00
10
0.19
1.9
201
6/9/2001
3:00
11
0.45
4.95
228
6/9/2001
4:00
12
0.01
0.12
258
6/9/2001
5:00
13
0
0
272
6/9/2001
6:00
225
6/9/2001
7:00
189
6/9/2001
8:00
142
6/9/2001
9:00
119
6/9/2001
10:00
98
6/9/2001
11:00
66
6/9/2001
12:00
42
6/9/2001
13:00
27
6/9/2001
14:00
0
SUM
1.83
14.55
2309.6
2-27
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2.13 Plot the hydrograph for the storm data given in Problem 1.22 (flow rate vs. time).
Label the following:
(a) peak flow Q p ,
(b) time to peak t p (distance from center of mass of rainfall to peak flow),
(c) time of rise TR (distance from start of discharge to peak flow),
(d) time base TB (distance from start of discharge to end of discharge).
ANSWER:
(a) peak flow Q p =1380 cfs
(b) time to peak t p = 1hr
(c) time of rise TR = 1.25hrs
(d) time base TB = 2.75 hrs
PROBLEM 2.14
2-28
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Below you are provided with a 30-min UH and the rainfall and infiltration information for a
storm event. Using this information, develop a storm hydrograph.
Time (hr)
U (m3/s)
i (cm/hr)
f (cm/hr)
0
0
0
0
0.5
15
0.75
0.25
1
40
1.5
0.2
1.5
60
3.0
0.2
2
85
1.75
0.1
2.5
105
0.5
0.1
3
120
3.5
110
4
95
4.5
80
5
60
5.5
45
6
30
6.5
10
7
0
30-min UH
Q (m3/s)
150
100
50
0
0
2
4
6
8
Time (hr)
2-29
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SOLUTION 2.14
We first need to determine the net rainfall for these periods:
Time (hr)
Rain (cm)
Infiltration (cm)
Net Rainfall (cm)
0
0
0
0
0.5
0.375
0.125
0.25
1.0
0.75
0.1
0.65
1.5
1.5
0.1
1.4
2.0
0.875
0.05
0.825
2.5
0.25
0.05
0.2
Now hydrograph convolution:
Time (hr)
U (m3/s)
P1*UH
P2*UH
P3*UH
P4*UH
P5*UH
Sum
0
0
0
0.5
15
0
1
40
3.75
0
1.5
60
10
9.75
0
2
85
15
26
21
0
2.5
105
21.25
39
56
12.375
0
128.625
3
120
26.25
55.25
84
33
3
201.5
3.5
110
30
68.25
119
49.5
8
274.75
4
95
27.5
78
147
70.125
12
334.625
4.5
80
23.75
71.5
168
86.625
17
366.875
5
60
20
61.75
154
99
21
0
3.75
19.75
62
2-30
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355.75
5.5
45
15
52
133
90.75
24
314.75
6
30
11.25
39
112
78.375
22
262.625
6.5
10
7.5
29.25
84
66
19
205.75
7
0
2.5
19.5
63
49.5
16
150.5
0
6.5
42
37.125
12
97.625
0
14
24.75
9
47.75
0
8.25
6
14.25
0
2
2
0
0
7.5
8
8.5
9
9.5
Storm Hydrograph
400
350
300
Q (m3/s)
250
200
150
100
50
0
0
-50
1
2
3
4
5
6
7
8
9
10
Time (hr)
2.15 Using a program such as Excel, develop the S-curve for the given 30-min UH, and
2-31
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then develop the 15-min UH from the 30-min UH.
Time (hr)
0
0.25
0.5
0.75
1.0
1.25
1.5
U (cfs)
0
15
70.9
118.6
109.4
81.6
60.9
Time (hr)
1.75
2.0
2.25
2.5
2.75
3.0
3.25
U (cfs)
45.4
33.9
25.3
18.9
14.1
10.5
7.8
Time (hr)
3.5
3.75
4.0
4.25
4.5
4.75
5.0
5.25
U (cfs)
5.8
4.4
3.3
2.4
1.8
1.6
0.8
0
ANSWER:
Create the S using the 30-min unit hydrograph. Lag by 30 min until you reach 0
Time
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
5.25
5.5
5.75
6
U (cfs)
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
7.8
5.8
4.4
3.3
2.4
1.8
1.6
0.8
0
30 min lagged UH
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
7.8
5.8
4.4
3.3
2.4
1.8
1.6
0.8
0
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
7.8
5.8
4.4
3.3
2.4
1.8
1.6
0.8
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
7.8
5.8
4.4
3.3
2.4
1.8
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
7.8
5.8
4.4
3.3
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
7.8
5.8
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
14.1
10.5
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
25.3
18.9
0
15
70.9
118.6
109.4
81.6
60.9
45.4
33.9
0
15
70.9
118.6
109.4
81.6
60.9
0
15
70.9
118.6
109.4
0
15
70.9
2.15 contโ
2-32
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0
S-curve
0
15
70.9
133.6
180.3
215.2
241.2
260.6
275.1
285.9
294
300
304.5
307.8
310.3
312.2
313.6
314.6
315.4
316.2
316.2
316.2
316.2
316.2
316.2
Time
S-curve
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
0
15
70.9
133.6
180.3
215.2
241.2
260.6
275.1
285.9
294
300
304.5
307.8
310.3
312.2
313.6
314.6
315.4
316.2
316.2
S-Curve
lagged (15
min)
Difference
15 min UH
(D/D’)=2
0
15
70.9
133.6
180.3
215.2
241.2
260.6
275.1
285.9
294
300
304.5
307.8
310.3
312.2
313.6
314.6
315.4
316.2
0
15
55.9
62.7
46.7
34.9
26
19.4
14.5
10.8
8.1
6
4.5
3.3
2.5
1.9
1.4
1
0.8
0.8
0
0
30
111.8
125.4
93.4
69.8
52
38.8
29
21.6
16.2
12
9
6.6
5
3.8
2.8
2
1.6
1.6
0
2.15 contโ
2-33
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2-34
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
PROBLEM 2.16
(a) Convert the following 1-hr UH into a 2-hr UH using Excel. Show table with time and
ordinates as well as the plotted UH. Find the area of this watershed.
(b) Given the net rainfall hyetograph, compute the storm hydrograph using hydrograph
convolution and the unit hydrograph you develop
ed in part, this time assuming it is a 1-hr UH (a). Show a table with times and ordinates as well as
the plotted storm hydrograph.
1-hr UH:
Time (hr)
U (cfs)
0
0
1
100
2
300
3
400
4
325
5
250
6
175
7
100
8
50
9
0
2.16 contโ
2-35
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Net Rainfall:
Time (hr)
Rainfall
Intensity
(in./hr)
0.25
0.8
0.5
0.8
0.75
0.8
1.0
0.6
1.25
1.0
1.5
1.2
1.75
0.8
2.0
0.6
2.25
0.6
2.5
0.2
2.75
0.2
3.0
0.2
3.25
0.1
3.50
0.4
3.75
0.4
4.0
0.1
4.25
0.1
4.5
0.1
4.75
0.04
5
0.04
2-36
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SOLUTION 2.16
(a) Lag the 1-hr UH by a 1-hr increment. Then add the lagged hydrograph to the original and
multiply the resulting ordinate values by the ratio of D/Dโ where D is the original
duration and Dโ is the desired duration (1/2)
Time
(hr)
1-hr
lagged
UH’s
1-hr
UH
0
1
2
3
4
5
6
7
8
9
10
0
100
300
400
325
250
175
100
50
0
Sum
2 hr
UH
0
100
400
700
725
575
425
275
150
50
0
0
100
300
400
325
250
175
100
50
0
0
50
200
350
362.5
287.5
212.5
137.5
75
25
0
2 hr UH
400
350
Flow (cfs)
300
250
200
150
2 hr UH
100
50
0
0
2
4
6
8
10
12
Time (hr)
2.16 contโ
2-37
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The area of the watershed will be found as follows:
Add the flows of the 2-hr UH: 1700 cfs
Multiply by the incremental hour: (1700 cfs)*(1 hr) = 1700 cfs-hr
Now convert this to acre-in.
If they assumed that 1 cfs-hr = 1 acre-in then they should get 1700 ac-in.
If they assumed 1 acre-in = 1.008 cfs-hr then they should get 1686.51 ac-in.
Because this is a unit hydrograph and we are therefore dealing with 1 inch of rainfall, the area of
the watershed is 1700 acres (or 1686.51 acres)
(b)
First we need to determine the hourly rainfall values from the table of intensities.
The rainfall for 0.25, 0.5, 0.75 and 1.0 –> hour 1
1.25, 1.5, 1.75, and 2.0 –> hour 2, and so on. So the rainfall totals that will be used for
hydrograph convolution are
2.16 contโ
2-38
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Time (hr)
Rainfall (in)
1
0.75
2
0.9
3
0.3
4
0.25
5
0.07
Now we do hydrograph convolution:
Rainfall: [0.75, 0.9, 0.3, 0.25,
0.07]
Time (hr)
P1*U
0
0
1
50
0
2
150
60
0
3
250
180
20
0
4
262.5
300
60
16.67
0
639.17
5
225
315
99.99
50
4.67
694.67
6
150
270
105
83.33
14
622.33
7
112.5
180
90
87.5
23.33
493.33
8
50
135
60
75
24.5
344.5
9
25
60
45
50
21
201
10
0
30
20
37.5
14
101.5
0
9.99
16.67
10.5
37.17
0
8.33
4.67
13
11
12
P2*U
P3*U
P4*U
P5*U
SH
0
50
210
450
2-39
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13
0
14
2.33
2.33
0
0
Problem 3 Storm Hydrograph
800
700
Flow (cfs)
600
500
400
300
200
100
0
0
2
4
6
8
10
12
14
Time (hr)
2.17 Given the following 2-hr unit hydrograph, calculate the 1-hr unit hydrograph. Then
back calculate and find the 2-hr unit hydrograph to prove that the method of calculation is
accurate. Graph both unit hydrographs against time on the same plot.
Time (hr)
0
1
2
3
4
5
6
Flow (cfs)
0
25
125
250
400
500
450
Time (hr)
7
8
9
10
11
12
13
Flow (cfs)
350
300
225
150
100
25
0
2.17 contโ
2-40
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16
ANSWER:
Procedure to obtain the 1 hr hydrograph:
~ Lag the 2-hr unit hydrograph by 2-hr increments to obtain the S curve
~ Then lag the S-curve by the time of duration of the new unit hydrograph (in this case, 1
hr)
~ Multiply the resulting ordinate values by the ratio of D/D’ where D is the original
duration and D’ is the desired duration (2/1 =2)
Time
2-Hr UH
S-curve
Lagged S-curve
0
0
0
1
25
25
2
125
0
3
250
25
4
400
125
5
500
6
Difference (cfs)
1 Hr UH
0
0
0
25
50
125
25
100
200
275
125
150
300
0
525
275
250
500
250
25
775
525
250
500
450
400
125
0
975
775
200
400
7
350
500
250
25
1125
975
150
300
8
300
450
400
125
0
1275
1125
150
300
9
225
350
500
250
25
1350
1275
75
150
10
150
300
450
400
125
0
1425
1350
75
150
11
100
225
350
500
250
25
1450
1425
25
50
12
25
150
300
450
400
125
0
1450
1450
0
0
13
0
100
225
350
500
250
25
1450
1450
0
0
The same procedure is done to convert to the 2 hr UH.
2-41
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2.17 contโ
~Lag the 1-hr unit hydrograph by 1-hr increments to obtain the S curve
~Then lag the S-curve by the time of duration of the new unit hydrograph (in this case, 2
hr)
~Multiply the resulting ordinate values by the ratio of D/D’ where D is the original
duration and D’ is the desired duration (1/2 =.5)
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
1-Hr UH lagged
0
50
200
300
500
500
400
300
300
150
150
50
0
0
0
50
200
300
500
500
400
300
300
150
150
50
0
0
50
200
300
500
500
400
300
300
150
150
50
0
50
200
300
500
500
400
300
300
150
150
0
50
200
300
500
500
400
300
300
150
0
50
200
300
500
500
400
300
300
0
50
200
300
500
500
400
300
0
50
200
300
500
500
400
0
50
200
300
500
500
0
50
200
300
500
0
50
200
300
0
50
200
0
50
0
Scurve
0
50
250
550
1050
1550
1950
2250
2550
2700
2850
2900
2900
2900
Lagged
Scurve
0
50
250
550
1050
1550
1950
2250
2550
2700
2850
2900
Blue = 1-hr UH, Red = 2-hr UH.
2-42
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
Difference
(cfs)
0
50
250
500
800
1000
900
700
600
450
300
200
50
0
1 Hr
UH
0
25
125
250
400
500
450
350
300
225
150
100
25
0
2.18 Develop storm hydrographs from UHs of subareas 1 and 2 shown in figure P2-18
for the given rainfall and infiltration.
Fig P2-18
t
i
f
(hr)
(in./hr)
(in./hr)
1
0.5
0.4
2
1.1
0.3
3
3
0.2
4
0.9
0.1
Time (hr)
0
1
2
3
4
5
6
7
UH1 (cfs)
0
200
450
650
450
300
150
0
UH 2 (cfs)
0
150
300
500
350
250
125
100
8
9
50
0
2-43
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2.18 contโ
ANSWER:
Time
(hr)
1
2
3
4
i
(in./hr)
0.5
1.1
3
0.9
f
(in./hr)
0.4
0.3
0.2
0.1
Net Rainfall Intensity
(in/hr)
0.1
0.8
2.8
0.8
Subarea 1
Time
UH1
(hr)
P1*UH1 P2*UH1 P3*UH1 P4*UH1 Q1
(cfs)
(cfs)
(cfs)
(cfs)
(cfs)
0
0
0
0
1
200
20
0
2
450
45
160
0
3
650
65
360
560
0
985
4
450
45
520
1260
160
1985
5
300
30
360
1820
360
2570
6
150
15
240
1260
520
2035
7
0
0
120
840
360
1320
8
0
0
420
240
660
9
0
0
120
120
10
0
0
0
20
205
2-44
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2.18 contโ
Subarea 2
Time
UH2
(hr)
P1*UH2 P2*UH2 P3*UH2 P4*UH2 Q2
(cfs)
(cfs)
(cfs)
(cfs)
(cfs)
0
0
0
0
1
150
15
0
2
300
30
120
0
3
500
50
240
420
0
710
4
350
35
400
840
120
1395
5
250
25
280
1400
240
1945
6
125
12.5
200
980
400
1592.5
7
100
10
100
700
280
1090
8
50
5
80
350
200
635
9
0
0
40
280
100
420
10
0
0
140
80
220
11
0
0
40
40
12
0
0
0
15
150
2-45
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2.18 contโ
2.19 Develop a combined storm hydrograph at point A in the watershed (Fig. P2โ18) and
lag route (shift in time only) assuming that travel time from point A to B is exactly
2 hours.
ANSWER:
We add the ordinates Q1 and Q2 from the previous problem to develop the combined
storm hydrograph at point A.
We then account for a 2-hr lag between points A and B to create a combined hydrograph
for A and B.
2-46
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2.19 contโ
Time (hr)
Q1 (cfs)
Q2 (cfs)
QA (cfs)
0
0
0
0
1
20
15
35
2
225
165
390
0
3
1030
740
1770
35
4
2030
1430
3460
390
5
2570
1950
4520
1770
6
2000
1568
3568
3460
7
1290
1068
2358
4520
8
630
620
1250
3568
9
105
412.5
517.5
2358
10
0
210
210
1250
11
35
35
517.5
12
0
0
210
QAB (cfs)
13
35
14
0
2-47
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2.20 Develop a storm hydrograph for subarea 3 from the given UH, add to the combined
hydrograph from Problem 2.19, and produce a final storm hydrograph at the outlet
of the watershed, B.
Time (hr)
UH3 (cfs)
I (in/hr)
f (in/hr)
0
0
0.5
0.4
1
140
1.1
0.3
2
420
3
0.2
3
630
0.9
0.1
4
490
5
350
6
210
7
130
8
70
9
0
ANSWER:
The net rainfall intensity was obtained in problem 2.18
time (hr)
I (in/hr)
f (in/hr)
Net Rainfall Intensity (in/hr)
1
0.5
0.4
0.1
2
1.1
0.3
0.8
3
3
0.2
2.8
4
0.9
0.1
0.8
2-48
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2.20 contโ
Time (hr) UH3 (cfs) P1*UH3 P2*UH3 P3*UH3 P4*UH3 Q3
0
0
0
1
140
14
0
2
420
42
112
0
3
630
63
336
392
0
4
490
49
504
1176
112
5
350
35
392
1764
336
6
210
21
280
1372
504
7
130
13
168
980
392
8
70
7
104
588
280
9
0
0
56
364
168
10
0
196
104
11
0
56
12
0
0
14
154
791
1841
2527
2177
1553
979
588
300
56
0
2.20 contโ
2-49
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
Time (hr) Q3 (cfs) QAB (cfs) Qtotal (cfs)
0
0
0
1
14
0
14
2
154
0
154
3
791
35
826
4
1841
355
2196
5
2527
1695
4222
6
2177
3380
5557
7
1553
4515
6068
8
979
3627.5
4606.5
9
588
2410
2998
10
300
1295
1595
11
56
540
596
12
0
220
220
13
40
40
14
0
0
2-50
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.21 Redo Example 2โ6 if the watershed is soil type B in good cover forest land. How
does the forested area compare to the meadow UH?
ANSWER:
From Table 2.1 SCS curve number is found to be 55. Therefore,
S = (1000 / CN) โ 10 = (1000 / 55) โ 10 = 8.18 in.
Convert miles into feet.
L = (5 mi)(5280 ft/mi) = 26,400 ft.
The slope is 100 ft/mi, so
y = (100 ft/mi)(1 mi / 5280 ft)(100%) = 1.9%
Thus, time to peak is
๏ฉ ( L) 0.8 ( S ๏ซ 1) 0.7 ๏น ๏ฉ (26,400) 0.8 (8.18 ๏ซ 1) 0.7 ๏น
tp = ๏ช
๏บ = ๏ช
๏บ = 6.21 hr
1900 1.9
๏ป
๏ช๏ซ 1900 y ๏บ๏ป ๏ซ
With rainfall duration D = 2 hr, the time to rise is
TR = D/2 + tp = 2/2 + 6.21 = 7.21 hr
Peak flow is
Qp = (484A / TR) = (484*10 / 6.21) = 779.4 cfs
where A is the area of 10 mi2
To complete the graph it is necessary to know the time of fall B.
B = 1.67TR = 12.04
2-51
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2.21 contโ
Forested Area (Group B)
Meadow (Group D)
TR
7.2 hr
4.36 hr
B
11.84 hr
7.17 hr
Qp
672.2 cfs
1110 cfs
S
8.18
2.82 in.
PROBLEM 2.22
A developer has recently purchased a 7 mi2 area of land north of Houston. She is
interested in determining the impact that development will have on the hydrologic
response of the area. There is one main channel running through the area and the length
along it from the outlet to the divide is 6 miles. The average slope in this area is 100
ft/mi. To determine the comparison hydrographs, a storm with duration of 3 hours will be
used.
Pre-development, the area is 40% wooded (good condition), 30% good condition
meadow, and 30% cultivated land with conservation treatment. Post-development the
watershed will be 10% wooded (good condition), 5% good condition meadow, 30%
cultivated land with conservation treatment, 30% residential with an average lot size of
1/8 acre, and 25% commercial area. The area has 40% soil type A and 60% soil type C.
Using the SCS method of unit hydrograph construction and Excel to aid in your
calculations and graphing, plot the resulting triangular hydrographs on the same graph for
comparison and note the differences in a few short sentences. How does development
affect the hydrograph? What do the resulting hydrographs tell you about the effects of
high or low curve numbers?
2-52
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SOLUTION 2.22
Pre-Development
1. Determine the curve number of the area based on the soil type and land use. Table with values
given on page 76 of textbook.
Land Use
Soil Group
Fraction of Area
CN
Wooded (Good
Condition)
A
(0.4)*(0.4) = 0.16
25
Wooded (Good
Condition)
C
(0.6)*(0.4) = 0.24
70
Meadow (Good
Condition)
A
(0.4)*(0.3) = 0.12
30
Meadow (Good
Condition)
C
(0.6)*(0.3) = 0.18
71
Cultivated Land (w/
conservation)
A
(0.4)*(0.3) = 0.12
62
Cultivated Land (w/
conservation)
C
(0.6)*(0.3) = 0.18
78
Weighted pre-development curve number:
CNpre = (0.16)*(25) + (0.24)*(70) + (0.12)*(30) + (0.18)*(71) + (0.12)*(62) + (0.18)*(78) =
58.66
2. Develop the pre-development SCS hydrograph parameters
S = (1000/CNpre) โ 10 = (1000/58.66) โ 10 = 7.047 in
2.22 contโ
2-53
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L = 6 mi = (6 mi)*(5280 ft/mi) = 31,680 ft
y = 100 ft/mi = (100 ft/mi)*(1 mi/5280 ft)*(100) = 1.9 %
?? =
?0.8 (?+1)0.7
(31,680)0.8 (8.047)0.7
=
= 6.553 hr
1900โ?
1900โ1.9
?
3
?? = 2 + ?? = (2) + 6.553 = 8.053 hr
?? =
484โ?
484โ7
= 8.053 = 420.71 cfs
??
Vol = (7 mi2)*(5280 ft/mi)2*(ac/43,560 ft)*(1 in) = 4480 ac-in
?=
2โ???
2โ4480
โ ?? = 420.71 โ 8.053 = 13.244 hr
??
Post-Development
1. Determine the curve number of the area based on the soil type and land use. Table with values
given on page 76 of textbook.
2.22 contโ
2-54
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Land Use
Soil Group
Fraction of Area
CN
Wooded (Good
Condition)
A
(0.4)*(0.1) = 0.04
25
Wooded (Good
Condition)
C
(0.6)*(0.1) = 0.06
70
Meadow (Good
Condition)
A
(0.4)*(0.05) = 0.02
30
Meadow (Good
Condition)
C
(0.6)*(0.05) = 0.03
71
Cultivated Land (w/
conservation)
A
(0.4)*(0.3) = 0.12
62
Cultivated Land (w/
conservation)
C
(0.6)*(0.3) = 0.18
78
Residential (1/8 acre
lots)
A
(0.4)*(0.3) = 0.12
77
Residential (1/8 acre
lots)
C
(0.6)*(0.3) = 0.18
90
Commercial
A
(0.4)*(0.25) = 0.1
89
Commercial
C
(0.6)*(0.25) = 0.15
94
Weighted post-development curve number:
CNpost = (0.04)*(25) + (0.06)*(70) + (0.02)*(30) + (0.03)*(71) + (0.12)*(62) + (0.18)*(78) +
(0.12)*(77) + (0.18)*(90) + (0.1)*(89) + (0.15)*(94) = 77.85
2. Develop the post-development SCS hydrograph parameters
S = (1000/CNpost) โ 10 = (1000/77.85) โ 10 = 2.845 in
L = 6 mi = (6 mi)*(5280 ft/mi) = 31,680 ft
2-55
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2.22 contโ
y = 100 ft/mi = (100 ft/mi)*(1 mi/5280 ft)*(100) = 1.9 %
?? =
?0.8 (?+1)0.7
(31,680)0.8 (3.845)0.7
=
= 3.908 hr
1900โ?
1900โ1.9
?
3
?? = 2 + ?? = (2) + 3.908 = 5.40774 hr
?? =
484โ?
484โ7
= 5.40774 = 626.509 cfs
??
Vol = (7 mi2)*(5280 ft/mi)2*(ac/43,560 ft)*(1 in) = 4480 ac-in
?=
2โ???
2โ4480
โ ?? = 626.509 โ 5.40774 = 8.894 hr
??
Graphs
The graphs may be either hand-drawn or developed in EXCEL. To plot, each graph will need
three points, connected in straight lines.
Pre-Development points: (0 , 0) , (8.053 , 420.71) , (21.297 , 0)
Post-Development points: (0 , 0) , (5.40774 , 626.509) , (14.302 , 0)
Notes on the graph: Because of the development of the area and the changes in the curve number
that are associated with an increase in imperviousness and other factors, the peak flow in the
developed hydrograph occurs earlier and is also greater than in is pre-development. These graphs
demonstrate that a higher curve number will translate to higher peak runoff.
2.22 contโ
2-56
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Direct Comparison of Pre and Post
Development SCS Hydrographs
700
600
Flow (cfs)
500
400
300
Pre-Development
200
Post-Development
100
0
0
5
10
15
20
25
Time (hr)
2-57
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2.23 A small watershed has the characteristics given below. Find the peak discharge Qp,
the basin lag time tp, and the time base of the unit hydrograph Tb, using Snyderโs
method.
A = 150 mi.2; Ct = 1.70; L = 27 mi.; Lc = 15 mi.; Cp = 0.7
ANSWER:
tp = Ct(LLc)0.3 = (1.70)(27*15 mi.2)0.3 = 10.3 hr
Qp =
640(C p )( A)
tp
=
640(0.7)(150)
= 6524 cfs
10.3
Since this is a small watershed, Tb ๏ป 4tp = 41.2 hr
2-58
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2.24 For a 55 mi2 watershed with Ct = 2.2, L = 15 mi., Lc = 7 mi., and Cp = 0.5, find tp,
Qp, Tb, and D. Plot the resulting Snyder UH.
ANSWER:
tp = Ct(LLc)0.3 = 2.2(15*7)0.3 = 8.9 hr
Qp =
640(C p )( A)
tp
=
640(0.5)(55)
= 1978 cfs
8.9
Since this is a small watershed, Tb ๏ป 4tp = 35.6 hr
D = tp/5.5 = 1.6 hr
TR = tp + D/2 = 8.9 + 1.6/2 = 9.7 hr
W75 = 440(Qp/A)-1.08 = 440(1978/55)-1.08 = 9.2 hr
W50 = 770(Qp/A)-1.08 = 770(1978/55)-1.08 = 16.1 hr
It is necessary to obtain more flow points in order to be able to draw the hydrograph. We
are aware the widths are distributed 1/3 before Qp and 2/3 after Qp so,
W75
*(1/3)
*(2/3)
W50
3.1
6.1
5.4
10.7
For W75, Q75 = 0.75Qp = 1484 cfs
t75,1 = 9.7 โ W75(1/3) = 6.6 hr
t75,2 = 9.7 + W75(2/3) = 15.8 hr
For W50, Q50 = 0.50Qp = 989 cfs
t50,1 = 9.7 โ W50(1/3) = 4.3 hr
t50,2 = 9.7 + W50(2/3) = 20.4 hr
Therefore, the following points should be plotted:
Time
(hr)
Flow (cfs)
0
4.3
6.6
9.7
0
989
1484
1978
2-59
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15.8
20.4
35.6
1484
989
0
2.24 cont’
2-60
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2.25 Assume the watershed in Example 2โ5 has gone through extensive commercial and
industrial growth on the wooded area. Now 50% of the formerly wooded areas have
become urbanized, so of that portion, 40% is commercial and business and 60% is
fair condition lawn space. Assume the soil is 50% group B and 50% group C for all
areas. Using Figure 2โ8, determine the runoff volume for a rainfall of 6 in.
ANSWER:
Originally, the watershed was 40% wooded. The new watershed is now 20% wooded
since 50% of the wooded area got developed. The other 20% is divided in the following
way:
Commercial and Business: 0.4*0.2 = 0.08
Fair Condition Lawn Space: 0.6*0.2 = 0.12
Land Use
Soil
Group
Fraction of Area
CN
Wooded
B
0.2*0.5 = 0.1
55
C
0.2*0.5 = 0.1
70
B
0.08*0.5 = 0.04
92
C
0.08*0.5 = 0.04
94
B
0.12*0.5 = 0.06
69
C
0.12*0.5 = 0.06
79
B
0.6*0.5 = 0.3
75
C
0.6*0.5 = 0.3
83
Commercial and Business
Fair Condition Lawn Space
Residential
Weighted CN:
CN = 55(0.1) + 70(0.1) + 92(0.04) + 94(0.04) + 69(0.06) + 79(0.06) + 75(0.3) + 83(0.3)
= 76
From Fig. 2-8 we get 3.4 in. direct runoff from 6 in. of rainfall.
2-61
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2.26 Make sure the unit hydrograph for Subbasin C is a unit hydrograph (Subbasin C
area = 770 ac). The unit hydrograph for Subbasin C is graphed in figure P2-27.
Subbasin C (Unit Hydrograph)
280
300.0
Q (cfs)
250.0
200.0
150.0
100.0
50.0
0.0
0
2
4
6
8
10
12
Time (0.5 hr)
Fig 2-27
ANSWER:
Area of Subbasin C = 770 ac
Qp = 280 cfs
DRO = Area under the hydrograph
DRO = (280 cfs)(5.5 hr)(1/2) = 770 cfs-hr = 770 ac-in.
770 ac-in./770 ac = 1 in.
This is a unit hydrograph.
2-62
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2.27
Using Fig 2-14, find the daily evaporation from a shallow lake with the following
characteristics:
Mean daily temperature = 25.6ยฐC,
Daily solar radiation = 550 cal/cm2,
Mean daily dew point = 4.4ยฐC,
Wind movement (6 in. above pan) = 5.5 ft/s.
ANSWER:
The chart in Fig. 2-14 is with different units of หF, cal/cm 2 , mi/day, and in. The given
measurements given need to be converted to the proper units
25.6ยฐC = (25.6)(1.8) + 32 = 78ยฐF
4.4ยฐC = (4.4)(1.8) + 32 = 40ยฐF
5. 5
ft 3600s 24hr 1mi.
mi.
ft
๏
๏
= 5.5 ๏
= 90
s
s
hr
day 5280 ft
day
Draw a bullet point at the intersection of T = 78ยฐF and radiation = 550 cal/cm2 (upper
left quadrant). Project a horizontal line to Td = 40ยฐF (upper right quadrant) and draw a
second point. Next project a vertical line to v = 90 mi./day (bottom right quadrant) and
draw a third point. Then, project a vertical line down from the first point and a horizontal
line to the left from the third point. The intersection of these two lines will be the point of
daily evaporation.
E = 0.25 in.
2-63
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2.28 A class A pan is maintained near a small lake to determine daily evaporation (see
table). The level in the pan is observed at the end of everyday. Water is added if the
level falls near 7 in. For each day the difference in height level is calculated
between the current and previous day, and the precipitation value is from the
current day. Determine the daily lake evaporation if the pan coefficient is 0.70.
Rainfall
Water Level
(in.)
(in.)
1
0
8.00
2
0.23
7.92
3
0.56
7.87
4
0.05
7.85
5
0.01
7.76
6
0
7.58
7
0.02
7.43
8
0.01
7.32
9
0
7.25
10
0
7.19
11
0
7.08*
12
0.01
7.91
13
0
7.86
14
0.02
7.8
Day
*Refilled at this point to 8 inches
ANSWER:
Pan evaporation = Change in water level in the pan + Amount of Rainfall
Lake Evaporation = (Pan evaporation)(coefficient)
2-64
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2.28 contโ
Rainfall
(in)
Day
Water Level (in.)
Pan Evaporation (in)
Lake Evaporation
(in)
1
0
8
0
0
2
0.23
7.92
0.31
0.217
3
0.56
7.87
0.61
0.427
4
0.05
7.85
0.07
0.049
5
0.01
7.76
0.1
0.070
6
0
7.58
0.18
0.126
7
0.02
7.43
0.17
0.119
8
0.01
7.32
0.12
0.084
9
0
7.25
0.07
0.049
10
0
7.19
0.06
0.042
11
0
7.08*
0.11
0.077
12
0.01
7.91
0.10*
0.070
13
0
7.86
0.05
0.035
14
0.02
7.8
0.08
0.056
* On Day 11, the pan is refilled. Thus, the pan evaporation that occurs on Day 12 (before
the measurement is taken) starts from 8 in, not 7.08 in.
So, 8.00 โ 7.91 = 0.09 in. = โ water level
Pan evaporation = โ water level + amount of rainfall = 0.09 in. + 0.01 in. = 0.10 in.
2-65
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2.29 Given an initial rate of infiltration equal to 1.4 in./hr and a final capacity of 0.6
in./hr, use Hortonโs equation [Eq. (2โ42)] to find the infiltration capacity at the
following times: t ๏ฝ 10 min, 15 min, 30 min, 1 hr, 2 hr, 4 hr, and 6 hr. You may
assume a time constant k = 0.24/hr.
ANSWER:
Hortonโs equation: f = fc + (fo โ fc)e-kt
f = 0.6 in./hr + (1.4 – 0.6 in./hr) e ๏ญ0.24t
f = 0.6 + .8 e ๏ญ0.24t plugging in given times (have to be in hours)
t (hr)
f(in/hr)
1
6
1.37
ยผ
1.35
ยฝ
1.31
1
1.22
2
1.10
4
0.91
6
0.79
2-66
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2.30 Determine a Horton equation to fit the following times and infiltration capacities.
Time
f
(hr)
(in./hr)
1
6.34
2
5.20
6.5
2.50
๏ฅ
1.20
ANSWER:
Hortonโs Equation: f = fc + (fo โ fc)e-kt
At t = โ,
f = 1.20 = fc + 0
fc = 1.20 (in./hr)
At t = 1 hr,
f = 6.34 = 1.20 + (fo โ 1.20)e-k
At t = 2 hr,
f = 5.20 = 1.20 + (fo โ 1.20)e-2k
Solve for f0 from t = 1 hr equation:
fo =
6.34 ๏ญ 1.20
๏ซ 1.20
e ๏ญk
Substitute into the t = 2 hr equation
f ๏ฝ 5.20 ๏ฝ 1.20 ๏ซ ((
5.20 = 1.20 ๏ซ
5.14
๏ซ 1.20) ๏ญ 1.20)e ๏ญ 2 k
๏ญk
e
5.14 ๏ญ 2 k
๏e
e ๏ญk
2.30 contโ
2-67
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
5.20 = 1.20 ๏ซ 5.14e ๏ญ k
4.00 = 5. 14e ๏ญ k
e ๏ญ k = 0.78
k = 0.25
Solve for f0
f = 6.34 = 1.20 + (fo โ 1.20)e-0.25
5.14 = (fo โ 1.20)e-0.25
fo = 7.8 in./hr
Plug known values into original equation:
f = 6.34 = 1.20 + (7.8 โ 1.20)e-0.25t
2-68
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2.31 A 5-hr storm over a 15-ac basin produces a 5-in. rainfall: 1.2 in./hr for the first
hour, 2.1 in./hr for the second hour, 0.9 in./hr for the third hour, and 0.4 in./hr for
the last two hours. Determine the infiltration that would result from the Horton
model with k = 1.1/hr, fc = 0.2 in./hr, and fo = 0.9 in./hr. Plot the overland flow for
this condition in in./hr vs. t.
ANSWER:
Time
Hortonโs Equation: f = fc + (f0 โ fc)e-kt
Infiltration
Rainfall
0
0.90
f = 0.2 + (0.9 โ 0.2)e-1.1t
1
0.43
1.2
f = 0.2 + 0.7e-1.1t
2
0.28
2.1
3
0.23
0.9
4
0.21
0.4
5
0.20
0.4
Plug in the values of t to solve for f.
To determine the volume of infiltration, take the integral of Hortonโs equation:
5
๏ญ1.1t
๏ฒ (0.2 ๏ซ 0.7e )dt
0
0.2๏t ๏0 ๏ซ
5
=
๏
๏
0.7 ๏ญ1.1t 5
e
0
๏ญ 1.1
= 1 ๏ซ (๏ญ0.636) ๏ (0.004 ๏ญ 1)
=1.633 in. over the watershed
2-69
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2.32 A plot of the infiltration curve obtained using Hortonโs equation is shown in Fig.
P2โ33. Prove that k ๏ฝ
( fo ๏ญ fc )
if Fยด is the area between the curve and the fc line.
Fยด
Find the area by integration over time, as time approaches infinity.
Fig 2.33
ANSWER:
A point is chosen on the curve such that at t = t 1 , f ๏ป fc. The area Fยด is equal to the area
under the curve above the line fc from t = 0 to t = to . This area may be found by the
integration as follows:
Let โtaโ be some arbitrary time.
ta
ta
F ‘ ๏ฝ ๏ฒ [ f c ๏ซ ( f o ๏ญ f c )e ]dt ๏ญ ๏ฒ f c dt
0
๏ญ kt
0
ta
๏ฝ ๏ฒ ( f o ๏ญ f c )e ๏ญkt dt
0
๏ฝ ( f o ๏ญ f c )(๏ญ1 / k )(e ๏ญ kta ) ๏ญ ( f o ๏ญ f c )(๏ญ1 / k )
As ta ๏ โ, e-kta ๏ 0. Therefore,
2-70
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2.32 contโ
F ‘ ๏ฝ (1 / k )( f o ๏ญ f c )
k ๏ฝ ( fo ๏ญ fc ) / F ‘
2.33 Determine the ๏ฆ index of Figure P2โ34 if the runoff depth was a)5.6 in. of rainfall
over the watershed area and b) 6.5 in.
ANSWER:
a. This is trial and error problem. From looking at the graph it is possible to set up the
equation:
(0.4 โ ฯ)6 + (0.5 โ ฯ)2 + (0.7 โ ฯ)4 + (0.9 โ ฯ)2 + (0.3 โ ฯ)4 + (0.2 โ ฯ)2 = 5.6
Assume ฯ = 0.2 in./hr,
(0.4 โ 0.2)6 + (0.5 โ 0.2)2 + (0.7 โ 0.2)4 + (0.9 โ 0.2)2 + (0.3 โ 0.2)4 + (0.2 โ
0.2)2 = 5.6
Both sides of the equation agree, so
2-71
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2.33 contโ
So ฯ index = 0.2 in./hr.
b. Again, this is trial and error problem. From looking at the graph it is possible to set up
the equation:
(0.4 โ ฯ)6 + (0.5 โ ฯ)2 + (0.7 โ ฯ)4 + (0.9 โ ฯ)2 + (0.3 โ ฯ)4 + (0.2 โ ฯ)2 = 6.5
With ฯ index = 0.1 in./hr = 7.1 so the index needs to be between 0.2 and 0.1
Trying ฯ index =.15 the answer is 6.6
with index = .155 the answer is 6.5 so the ฯ index for a rainfall of 6.5 in. is .155 in./hr.
2.34
A sandy loam has an initial moisture content of 0.18, hydraulic conductivity of 7.8
mm/hr, and average capillary suction of 100 mm. Rain falls at 2.9 cm/hr, and the
final moisture content is measured to be 0.45. When does surface saturation occur?
Plot the infiltration rate vs. the infiltration volume, using the Green and Ampt
method of infiltration.
ANSWER
ฮธi = 0.18
ฮธf = 0.45
Ks = 78 mm/hr = .78 cm/hr
ฮจ = โ100 mm = -10 cm
i = 2.9 cm/hr (for 6 hours)
Md = ฮธf – ฮธi = 0.45 – 0.18 = 0.27
FS ๏ฝ
๏๏MD
๏ญ 100mm ๏ 0.27
= 9.93 mm
๏ฝ
๏ฆ i ๏ถ
๏ฆ 2.9cm / hr ๏ถ
๏ท๏ท 1 ๏ญ ๏ง
๏ท
1 ๏ญ ๏ง๏ง
๏จ 0.78cm / hr ๏ธ
๏จ KS ๏ธ
Until 9.93 mm has infiltrated, the rate of infiltration is equal to the rainfall rate (2.9
cm/hr). This will happen at:
2-72
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.33 contโ
.993 cm ๏
1hr
= 0.34hr
2.9cm
After surface saturation, the rate of infiltration is calculated using:
๏ฆ M ๏๏ถ
f ๏ฝ K S ๏ง1 ๏ญ d ๏ท
F ๏ธ
๏จ
F (cm)
f (cm/hr)
1
2
3
4
5
6
7
8
2.89
1.83
1.48
1.31
1.20
1.13
1.08
1.04
2-73
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.35 Use the parameters given to graph the infiltration rate vs. the infiltration volume for
the same storm for both types of soil. Prepare a graph using the GreenโAmpt
method, comparing all the curves calculated with both the lower-bound and upperbound porosity parameters. The rainfall intensity of the storm was 1.5 cm/hr for
several hours, and the initial moisture content of all the soils was 0.15.
Capillary Suction
Hydraulic Conductivity
Soil
Porosity
(cm)
(cm/hr)
Silt loam
0.42 โ 0.58
16.75
0.65
Sandy clay
0.37 โ 0.49
23.95
0.10
ANSWER:
ฮธi = 0.15
ฮจ = – capillary suction
i = 1.5 cm/hr
Md = ฮธ f – ฮธ i
SL
0.65
SC
0.10
ฮจ (cm)
-16.75
-23.95
ฮธf
0.42 to 0.58
0.37 to 0.49
K s (cm/hr)
Silty Loam (SL):
FS ๏ฝ
๏ญ 16.75cm ๏ M d
๏๏MD
๏ฝ
๏ฝ 12.81M d
๏ฆ i ๏ถ
๏ฆ 1.5cm / hr ๏ถ
๏ท๏ท 1 ๏ญ ๏ง
๏ท
1 ๏ญ ๏ง๏ง
๏จ 0.65cm / hr ๏ธ
๏จ KS ๏ธ
For low porosity:
Md = 0.42 – 0.15 = 0.27
Fs = 3.46 cm
Saturation time = Fs/i = 2.31 h
2-74
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.35 contโ
For high porosity:
Md = 0.58 – 0.15 = 0.43
Fs = 5.51 cm
Saturation time = Fs/i = 3.67 hr
Sandy Clay (SC):
FS ๏ฝ
๏ญ 23.95cm ๏ M d
๏๏MD
๏ฝ
๏ฝ 1.71M d
๏ฆ i ๏ถ
๏ฆ 1.5cm / hr ๏ถ
๏ท๏ท 1 ๏ญ ๏ง
๏ท
1 ๏ญ ๏ง๏ง
๏จ 0.10cm / hr ๏ธ
๏จ KS ๏ธ
For low porosity:
Md = 0.37 – 0.15 = 0.22
Fs = .38 cm
Saturation time = Fs/i = 0.25 hr
For high porosity:
Md = 0.49 – 0.15 = .34
Fs = .58 cm
Saturation time = Fs/i = 0.39 hr
๏ฆ ๏M d ๏ถ
Plot the infiltration volume vs. the infiltration rate using: f ๏ฝ K S ๏ง1 ๏ญ
๏ท
F ๏ธ
๏จ
SL
SC
f lp
F
0.25
0.5
1
2
3
4
5
6
7
8
9
10
3.6
3.6
1.6
1.4
1.2
1.1
1.1
1.0
1.0
0.9
f hp
5.3
3.0
2.2
1.8
1.6
1.4
1.3
1.2
1.2
1.1
f lp graph
f hp graph
1.5
1.5
1.5
1.5
1.5
1.4
1.2
1.1
1.1
1.0
1.0
0.9
f lp
F
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.4
1.3
1.2
1.2
1.1
0.25
0.5
1
2
3
4
5
6
7
8
9
10
2.2
1.2
0.6
0.4
0.3
0.2
0.2
0.2
0.2
0.2
0.2
0.2
f hp
f lp graph
3.4
1.7
0.9
0.5
0.4
0.3
0.3
0.2
0.2
0.2
0.2
0.2
2-75
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
f hp graph
1.5
1.2
0.6
0.4
0.3
0.2
0.2
0.2
0.2
0.2
0.2
0.2
1.5
1.5
0.9
0.5
0.4
0.3
0.3
0.2
0.2
0.2
0.2
0.2
2.35 contโ
2-76
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.36 The Green and Ampt infiltration equation is a loss function used to compute the
cumulative infiltration, F (cm) for a given infiltration rate, f (cm/hr). Recall that,
f=Ksยท(1-(Mยทwf )/F). For the given soil properties and infiltration rate, answer the
following.
K
1
cm/hr
wf
-11
cm
M
0.2
t(hr)
f(cm/hr)
F(cm)
0
0
0
0.01
15.34
0.15
0.25
3.32
0.95
0.50
2.42
1.55
0.75
2.06
2.07
a. Compute the infiltration rate, f (cm/hr) for F=2.07 cm, and show your computations.
f=Ksยท(1-(Mยทwf )/F) = 1(1-0.2*(-11)/2.07) =2.06 cm/hr
b.
Compute the cumulative infiltration resulting from a constant rain rate of 0.5 cm/hr for 1
hr.
F=0.5 * 1 = 0.5 cm
c.
At saturation, what is the infiltration rate in cm/hr? Justify.
At saturation, f=K=1 cm/hr
2-77
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.37 Please refer to the Green and Ampt Infiltration and Runoff Example posted on the textbook
website along with the associated Excel Spreadsheet. Complete the problem in the
example, and then repeat the procedure with a constant rainfall rate of 50 mm/hr for 1.8
hrs. Determine the new cumulative runoff and runoff coefficient.
Infiltration Rate, f (mm/hr)
ANSWER:
Time, t (hr)
Rainfall (mm/hr)
Infiltration Rate (mm/hr)
Ks =
wf =
Md =
6.5 mm/hr
-166.8 mm
0.75 Moisture Deficit
Cumulative Rainfall (mm) =
Cumulative Runoff (mm) =
Cumulative Infiltration
(mm) =
Runoff Coefficient =
92.5
32.1
60.4
34.7%
78
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.38 A constant rain rate storm (2 cm/hr for 4 hr) falls on a silt loam soil with an effective
saturation of 30%. Determine the infiltration rate (f) and cumulative infiltration (F) at time t
= 1 hr. Determine the time (t) and infiltration rate (f) when the cumulative infiltration F = 6
cm.
Solution
Values obtained from problem statement and table 2-4 for silt loam.
?? =
?? โ (? โ ?? )
, ?? ?? = ?? โ (? โ ?? )
??
0.3 โ 0.486 = ?? โ (0.501 โ 0.486),
?? = 0.1608
?? = ?? (1 โ ?? ) = 0.486(1 โ 0.3) = 0.3402
? = ?? (?? โ
? โฯ
?
?? = 1โ?/?
=
ฯ
0.3402
3.688
+ 1) = 0.65 (16.68 โ
+ 1) =
+ 0.65
F
?
?
0.3402โโ16.68
1โ2/0.65
= 2.732??
??
= 1.366 โ?
2
a) When t=1, f= 2cm/hr, F= i*t= 2 cm/hr
?? =
b) When F= 6cm, ? =
3.688
6
+ 0.65 = 1.265 ??/โ?
ฯ โ ?? โ ?
?? (? โ ?? ) = ? โ ?? + ฯ โ ?? โ ln (
)
ฯ โ ?? โ ??
โ16.68 โ 0.3402 โ 6
0.65(? โ 1.366) = 6 โ 2.732 โ 16.68 โ 0.3402 โ ln (
)
16.68 โ 0.3402 โ 2.732
? = 3.527โ?
79
ยฉ 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved.
2.39
Repeat Example 2โ11 in the chapter with the same rainfall, except use Hortonโs equation
for the infiltration with f0 = 1.2 in/hr, fc = 0.2 in/hr, and k = 0.35hrโ1.
Solution
(Assuming the runoff is unknown, and we are trying to find it.)
f0
fc
k
A
1.2
0.2
0.35
0.875
in/hr
in/hr
1/hr
mi^2
First, put the area into units of acres: 0.875 mi2 * 640 ac/mi2 = 560 acres
Then add the rainfall values for each time period to get total rainfall:
time
i
0
1
2
3
4
5
6
7
8
9
10
11
12
0
1.4
1.4
2.3
2.3
2.3
1.1
1.1
0.7
0.7
0.7
0.3
0.3
14.6 in total rainfall
Then use equation (2-43) with the given parameters to solve for the total infiltration:
? โ?
?(?) = ?? โ ? + 0 ? ? โ (1 โ ? โ?? ),
1.2 โ 0.2
โ (1 โ ? โ0.3โ12 ) = 5.21 in infiltrated
0.35
Then subtract the rain infiltrated from the rain fallen
?(?) = 0.2 โ 12 +
14.6-5.21 = 9.29 in, 9.29 in*1ft/12 in = 0.78 ft
Multiply this amount of rain by the area of the watershed to get total runoff
0.78 ft*560 acres = 438 acre-ft runoff
80
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