Lafayette College
Department of Civil and Environmental Engineering
CE 321: Introduction to Environmental Engineering
Fall 2010
SOLUTIONS - Homework #2 – SOLUTIONS
Due: Friday, 9/13/10
SOLUTIONS
1. A watershed has an area of 25 square miles. A rain event of 0.25 inch falls on the watershed.
One-third of the rain infiltrates into the ground, and the rest becomes runoff.
a. What is the total volume of rain? (report your answer in acre-ft)
b. What is the total volume of runoff? (report your answer in acre-ft)
c. If some of the vegetation in the watershed is removed and replaced with a housing
development, would the runoff increase or decrease? Explain your answer?
A watershed has an area of 25 square miles. A rain event of 0.25 inches falls on the watershed.
One third of the rain infiltrates into the ground, and the rest becomes runoff.
Rainfall = 0.25 inches
Watershed =
25 square miles
Runoff = 1 – 1/3
Infiltration = 1/3 Rainfall
a. What is the total volume of rain? (report your answer in acre-ft)
Vol = 0.25 in (1ft/in) (25 mi2 x(5280 ft/mi)2) = 1.45 x 107 ft3
1.45 x 107 ft3 / (43560 ft3/acre-ft) = 332.87 acre-ft
b. What is the total volume of runoff? (report your answer in acre-ft)
Since 1/3 infiltrates, then 2/3rds must become runoff.
Vol = 2/3 (1.45 x 107 ft3) = 9.68 x 106 ft3
9.68 x 106 ft3 / (43560 ft3/acre-ft) = 222.22 acre-ft
c. If some of the vegetation in the watershed is removed and replaced with a housing
development, would the runoff increase or decrease? Explain your answer?
Runoff would increase due to increased impervious areas (driveways, rooftops),
and decreased vegetation (which promotes infiltration and evaporation).
2. Find a maximum reservoir storage requirement (MG/month) if a uniform draft of 726,000 gpd/mi2 from a specific stream is to be maintained.
Perform both a graphical and tabular analysis. Explain any differences in your final storage requirement between the two methods. The following
record of average monthly runoff values is given:
Mo*
A
M
J
J
A
S
O
N
D
J
F
M
A
M
J
RO**
97
136
59
14
6
5
3
7
19
13
74
96
37
63
49
*Mo = Month, **RO = Runoff (mgal/mi2/mo)
DRAFT = 726,000 gpd/mi^2 (flow to meet water demand)
0.726
28 day month
30 day month
31 day month
28
30
31
2419200
2592000
2678400
sec/28 day month
sec/30 day month
sec/31 day month
Month
Inflow (I)
(MG/mi^2/mo)
OutFlow (O)
(Draft)
(MG/mi^2/d)
OutFlow (O)
(Draft)
(MG/mi^2/mo)
Cumulative
Inflow (MG)
(Graphical)
Cumulative
Ouflow
(MG)
(Graphical)
Deficiency
I-O
(MG/mi^2/mo
Cumulative
Deficiency
(MG/mi^2/mo)
April
May (31)
June
July (31)
Aug (31)
Sept
Oct (31)
Nov
Dec (31)
97.00
136.00
59.00
14.00
6.00
5.00
3.00
7.00
19.00
7.26E-01
7.26E-01
7.26E-01
7.26E-01
7.26E-01
7.26E-01
7.26E-01
7.26E-01
7.26E-01
2.18E+01
2.25E+01
2.18E+01
2.25E+01
2.25E+01
2.18E+01
2.25E+01
2.18E+01
2.25E+01
9.70E+01
2.33E+02
2.92E+02
3.06E+02
3.12E+02
3.17E+02
3.20E+02
3.27E+02
3.46E+02
2.18E+01
4.43E+01
6.61E+01
8.86E+01
1.11E+02
1.33E+02
1.55E+02
1.77E+02
2.00E+02
75.22
113.49
37.22
-8.51
-16.51
-16.78
-19.51
-14.78
-3.51
0.00
0.00
0.00
-8.51
-25.01
-41.79
-61.30
-76.08
-79.58
Jan (31)
13.00
7.26E-01
2.25E+01
3.59E+02
2.22E+02
-9.51
-89.09
Feb (28)
March (31)
April
May (31)
June
74.00
96.00
37.00
63.00
49.00
7.26E-01
7.26E-01
7.26E-01
7.26E-01
7.26E-01
2.03E+01
2.25E+01
2.18E+01
2.25E+01
2.18E+01
4.33E+02
5.29E+02
5.66E+02
6.29E+02
6.78E+02
2.42E+02
2.65E+02
2.87E+02
3.09E+02
3.31E+02
53.67
73.49
15.22
40.49
27.22
-35.42
0.00
0.00
0.00
0.00
3. A 4,000-km2 watershed receives 102 cm of precipitation in one year. The average flow of
the river draining the watershed is 43.2 m3/s. Infiltration is estimated to be 5.5 X 10-7 cm/s
and evapo-transpiration is estimated to be 40 cm/y. Determine the change in storage in the
watershed over one year. (report your answer as m3) The ratio of runoff (in cm) to
precipitation is termed the runoff coefficient. Computer the runoff coefficient for this
watershed.
Evaporation = 40 cm/yr
Precipitation = 102 cm
Watershed = 4,000 km2
Outflow = 43.2 m3/s
Infiltration = 5.5 x 10-7 cm/y
Known:
Precipitation = P = 102 cm/y
Evaporation = E = 40 cm/y
Watershed Area = 4,000 km2
Infiltration = I = 5.5 x 10-7 cm/y
Runoff = Outflow = O = 43.2 m3/s
Find:
a) Change in Storage over one year = ∆S/y
b) Ratio of runoff (cm) to perception = runoff coefficient
Assumption:
No additional reactions, Water activity is as defined…no other source or removals, constant
temp.
Solution:
∆S/y = P – E – I – O
Runoff = O =
I=
= 34.05 cm/y
= 17.34 cm/y -
a)
∆S/y = 102 cm/y - 40 cm/y - 40 cm/y - 34.05 cm/y = 10.61 cm/y
OR
∆S for one year reported as Volume (m3) considering a watershed area of 4,000 km2
= 4.2 x 108 m3
Volume =
b) Runoff coefficent
=
= 0.3333 or 33.33%