Homework 3 answers
Note: These answers do not strictly display significant figures.
7
7.25 A nonhomogeneous watershed has the following distribution of landcover:
Fraction
0.25
0.30
0.20
0.10
0.15
Landcover
Cultivated land
Forest
Pasture
Open space
Open space
Soil Group
C
D
C
B
C
Slope
(%)
2.3
2.2
1.9
1.8
2.1
Runoff
Coefficient from
Table 7.9
0.19
0.16
0.24
0.08
0.17
(Note: I added the last column.)
Estimate the runoff coefficient for use in predicting a 10-yr peak discharge.
Cw 
W C
W
i
i
i

0.25 * 0.19  0.30 * 0.16  0.20 * 0.24  0.10 * 0.08  0.15 * 0.17
 0.177
0.25  0.30  0.20  0.10  0.15
7.27 Using the Rational Method, estimate the 10-yr peak discharge from a 35-acre
watershed using the IDF curve shown in Figure 4.4. Assume light industrial
development with the principal flowpath being paved surfaces including gutter flow.
The principal flowpath has a length of 1,800 ft and a slope of 2%.
Table 7.10 gives a recommend runoff coefficient of 0.65 for light industrial areas.
Table 3.14 gives k = 9.1 for a paved surface, thus velocity is:
v  kS 0.5  9.1 * 0.02 0.5  1.28 ft / s
tc 
L

v60
1,800 ft
 23 min
ft
s
1.28 * 60
s
min
For a duration of 23 minutes and a return period of 10 years, Fig. 4.4 gives intensity
of 4 inches per hour. So:
q p  CiA  0.65 * 4 * 35  91 ft 3 / s
Alternatively, Kirpich’s formula gives tc=11 min. Using that duration, i=6 in/hr, so
qp=137 ft3/s. It is also reasonable to say that k = 42. Then v = 6 ft/s and tc=5
minutes. For this short duration, i=7.9 in/hr, so qp = 180 ft3/s.
Homework 3 Answers, Page 2 of 3
7.43 Compute the weighted runoff curve number for the following conditions and use
it to compute the depth of runoff (in.) for a storm depth of 4.5 in.
Area
Fraction
Land Use/Condition
0.30
Lawns, open space/fair
0.25
Residential/0.5-acre lots
0.45
Woods/good
Soil
CN from
Group Table 3.18
C
79
D
85
C
70
(Note: Last column added.)
CN av 
W CN
W
i

i
i
0.30 * 79  0.25 * 85  0.45 * 70
 76.45
0.30  0.25  0.45
1000
1000
 10 
 10  3.08in; I A  0.2 * S  0.2 * 3.08 in  0.62 in
CN
76.45
 P  I A 2
4.5  0.622  2.17 in
Q

P  I A  S 4.5  0.62  3.08
S
7.45 Compute the curve number for a 20-acre drainage area that has the following
characteristics. Compute the depth of runoff (in.) for a storm of 5 in. Compute the
runoff using both a weighted CN approach and a weighted Q approach
Area
(acres)
Imperviousness
%
Soil
Group
CNp from p. 165 for
use in Eq’n 3.50
4
6
7
3
24
18
60
75
B
C
B
C
61
74
61
74
CN w  CN p 1  f   f 98
69.88
78.32
83.2
92
S
(in.)
Q
(in.)
4.31
2.77
2.02
0.87
2.03
2.74
3.19
4.09
For the weighted CN approach:
CN av 
W CN
W
i
i
i

0.2 * 69.88  0.3 * 78.32  0.35 * 83.2  0.15 * 92
 80.39
0.1  0.3  0.35  0.15
1000
1000
 10 
 10  2.44 in
CN
80.39
2
2


P  0.2 S 
5  0.2 * 2.44 
Q

 2.93 in
P  0.8S
5  0.8 * 2.44
S
For the weighted Q approach:
Qav 
W Q
W
i
i
i

0.2 * 2.03  0.3 * 2.74  0.35 * 3.19  0.15 * 4.09
 2.96 in
0.1  0.3  0.35  0.15
Homework 3 Answers, Page 3 of 3
7.49 Assuming C soil and the watershed conditions of Problem 7.27, estimate the peak
discharge with the SCS Graphical Method.
To get qu, we need tc and IA/P. We will assume Type II rainfall since the IDF curve is
for Baltimore. (See Fig. 4.14 on p. 213.) To keep things simple, let’s use the same tc
we found in 7.27, 23 minutes is 0.38 hours. Table 3.18 gives 91 for the CN for
industrial districts on C soil. From Figure 4.4, a 24-hour, 10-yr storm has an intensity
of 0.21 in/hr, thus a total storm depth of 5.04 inches. So:
1000
1000
 10 
 10  0.99 in; I A  0.2 * S  0.2 * 0.99 in  0.20 in
CN
91
IA
0. 2

 0.04  Use 0.1
P 5.04
S
From Figure 7.9c, qu is about 610 ft3/(s·mi2·in). Pe is obtained as:
Pe
2

P  0.2S 

P  0.8S
2

5.04  0.2 * 0.99

5.04  0.8 * 0.99
 4.02 in
Area is 35 acres, which need to be converted to square miles. Since we don’t know
of any ponds or swamps within the watershed, we’ll use FP=1. Then:
q p  qu APe F  610
ft 3
1mi 2
ft 3
*
35
ac
*
*
4
.
02
in
*
1

134
640ac
s
s  mi 2  in
Alternatively, using tc = 11 min ≈ 0.2 hours, qu = 810 ft3/(s·mi2·in). So qp = 178 ft3/s.
Using tc = 5 minutes ≈ 0.1 hr gives qu off the top of the chart, about 1020
ft3/(s·mi2·in). So qp = 224 ft3/s