HW 1:
Homework, Tests & Quiz Solutions
Problem 1:
The chart from a gauge shown represents the record between 6 AM Sunday and Noon
Monday. Find the average rainfall intensity during this period. Also find the total
precipitation between the same period.
0.35
0.30
0.25”
0.20”
0.15
0.1”
0.05”
0
6 AM 9 AM Noon
Solution:
3 PM
6 PM
9 PM
12 PM 3 AM 6 AM 9 AM
12 Noon
Average rain= The area under the curve / Total time
= {[(0.05+0.20)/2](6hrs)+[0.2](3hrs)+[{0.2 +0.3)/2](6hrs)+[(0.3 +
0.35)/2](12hrs) + [0.35](3hrs)} (1/30hrs)
= (0.75+0.6+1.5+3.9+1.05)/30 = 0.26 “
Average intensity = 0.26/30 hrs = 0.09 inches/hr
Total rain = (Intensity) (duration) = 0.26”
Problem 2:
In a given year, a 1000 mi2 watershed received 12 inches of precipitation. The annual rate of
outflow measured in the river draining the area is 600 ft3/sec. Estimate the
Evapotranspiration. Assume negligible change of storage and no net groundwater flow.
Solution:
ET = P – O
ET = 12 – (600 ft3/s) (31,536,000 s/yr) (1/1000 mi2 x 52802 ft2/mi2) = 12 – 0.679 = 11.321 “
1
Problem 3:
Consider a reservoir with one inlet stream, one outlet at a dam and a surface area of .5 km 2.
The reservoir level after weeks of drought is falling at a rate of 3 mm/day. The average
evaporation rate from the reservoir surface is 1.2 mm/day, the inlet discharge is 10,000
m3/day, and the outlet discharge is 16,000 m3/day. Assuming the only variables in the
budget equation are: I,G,O,E (inflow, Groundwater, outflow, evaporation and rate change of
storage dV/dt where I + G – O –E = dV/dt. Find the total net rate of groundwater discharge
into the reservoir
Solution:
dQ / dt = -3 mm/d x 10-3 m/mm = - 0.003 m/d
E = 1.2 x 10-3 m/mm = 0.0012 m/d
I - O = 10,000 - 16,000- = - 6000 m3/d
I – O in m = (- 6000 m3/d) / (0.5 x 1000,000 m2) = - 0.012 m/d
I – O – E + G = dV/dt
- 0.012 – 0.0012 + G = - 0.003
G = + 0.012 + 0.0012 – 0.003 = 0.01 m
Problem 4:
Consider the following data:
Gage
A
B
39
X
Annual ppt
42
41
39
41
Storm Event
2.6
3.1
2.3
?
Find the missing data for station X
Solution:
Px = (Ax / n AA) PA + (Ax / n AB) PB + (Ax / n AC) PC + (Ax / n AD) PD
AX = 41
N=3
Px = [41/(3x42)] (2.6) + [41/(3x41)] (3.1) + [41/(3x39)] (2.3) = 0.846 + 1.033 + 0.806 = 2.685
2
HW 2:
3
4
5
HW No 3:
The infiltration rate for small area was observed to be 4.5 in/hr at the beginning of the rain,
and it decreased exponentially to an equilibrium of 0.5 in/hr after 10 hrs. a total of 30 inches
of water infiltrated during the 10 hr interval. Determine the value of k in Horton equation.
Solution:
f = f∞ + (f0 - f∞) e-Kt
f∞ = 0.5 in/ hr and ∫0 to 10 f (t) dt = 30 in
f = 4.5 in / hr then ∫0 to 10 [ f∞ + (f0 - f∞) e-Kt ] dt = 30
then: 10 f∞ + (f0 - f∞)/k [1-e-10K] = 30
Then: (10)(0.5) + (4.5-0.5) / k (1-e-10K) = 30
then K = 0.1027 hr-1
6
Given initial infiltration capacity f0 of 50 cm/day and a time constant k of 0.20 hr-1 derive
infiltration capacity curve vs. time if the final infiltration capacity is 10 cm/day. Estimate the
infiltrated water in m3 for the first 10 hours for 100 km2 watershed
Solution:
f0 = 50/24 = 2.083 cm/hr
f∞ = 10/24 = 0.417 cm/hr
k = 0.2 hr-1
The infiltration capacity curve is:
f = 0.417 + (1.666) e – 0.2 t
F = tf∞ + [(f0 - f∞)/k ] [1-e-0.2K] = (10 hr) (0.417 cm/hr) + (1.666)/0.2) [1 - e-0.2(10)]
= 4.17 + 8.33 (1-e-2) = 5.6 cm
Total infiltrated water: (100,000)2 (5.6 cm x 0.01 m/cm) = 560x106 m3
HW No. 5:
PROBLEM 1:
The following is the discharge rates resulting from 2 - hr unit hydrograph:
Time (hr):
Q (cfs):
0
0
2
100
4
250
6
200
8
100
10
50
12
0
1) Develop the 4-hr unit hydrograph.
2) Find the total runoff resulting from the following rain:
Time increment:
First 4 hrs
Second 4 hrs
Rain Intensity (inches/hr)
0.5
1.5
Solution:
Time, hr
(1)
2-hr UH
CFS
(2)
0
2
4
6
8
10
0
100
250
200
100
50
UHLagged
CFS
(3)
0
100
250
200
100
Σ
(2)+(3),
CFS
(4)
0
100
350
450
300
150
4-hr UH
(4) / 2,
CFS
(5)
0
50
175
225
150
75
4-hr UH x
(0.5)
CFS
(6)
0
25
87.5
112.5
75
37.5
4-hr UH x
(1.5)
CFS
(7)
0
75
262.5
337.5
225
Σ
CFS
(8)
0
25
162.5
375
412.5
262.5
7
12
14
0
50
0
50
0
25
0
12.5
0
112.5
37.5
0
125
37.5
0
Total Runoff:
(1/2) [(0+25)+(25+162.5)+(162.5+375)+(375+412.5)+(412.5+262.5)+(262.5+125)+(125+37.5)+
(37.5+0)[(2 hrs)(60x60) = (25+162.5+375+412.5+262.5+125+37.5) (7200) = 8.19x106 ft3
Total Runoff:
(1/2) [(0+25)+(25+162.5)+(162.5+375)+(375+412.5)+(412.5+262.5)+(262.5+125)+(125+37.5)+
(37.5+0)[(2 hrs)(60x60) = (25+162.5+375+412.5+262.5+125+37.5) (7200) = 8.19x106 ft3
PROBLEM 2:
Given the following 2-hr unit hydrograph, use the S-curve to develop the ordinates of 3-hr
unit hydrograph:
Time (hr):
Q (cfs):
Time
(hr)
(1)
0
0
Q
(cfs)
1
200
2
500
S-Curve
additions
(3)
3
400
4
200
5
100
6
0
SLagged Difference
Curve
SCurve
(6)
(4)
0
0
3-hr
UH =
(6) x
2/3
0
0
(2)
0
-
1
200
-
200
200
133.3
2
500
-
500
500
333.3
3
400
200
600
0
600
400
4
200
500
700
200
500
333.3
5
100
400+200
700
500
200
133.3
6
0
200+500
700
600
100
66.7
8
7
100+400+200
700
700
0
0
700
HW No. 6:
Problem 2:
The following field data was taken from piezometers installed side by side at a single
site (each 100 m apart):
Piezometer:
Elevation at surface (m)
Depth of Piezometer (m)
Depth to water (m)
A
450
150
27
B
450
100
47
C
450
50
36
If A, B and C (m) refer to the points of measurement of piezometers a, b and c.
Calculate:
a)
b)
c)
d)
The hydraulic head at A, B and C (m)
The pressure head at A, B and C (m)
The elevation head at A, B and C (m)
The hydraulic gradient between A and B and B and C.
A
B
450
100 m
C
100 m
423
414
403
400
350
300
Datum
9
a) Hydraulic head @ A = 423 - 300 = 123 m
B = 50 + (403-350) = 103 m
C = 100 + (414-400) = 114 m
b) Pressure head @ A: 423 – 300 = 123 m
B: 403 – 350 = 53 m
C: 414 – 400 = 14 m
a) Elevation heads: A= 0, B= 50 m and C= 100 m
b) Gradient A to B = (123-103) / 100 = 0.2
c) Gradient B to C = (103-114) / 100 = -0.11
Problem 3:
A soil sample is tested in laboratory. The column has an inside diameter of 10 cm and the
length of the soil sample is 25 cm. With steady flow of Q = 1.7 cm 3 / min, the head difference
between the two manometers is 15 cm. Calculate:
a) The hydraulic conductivity of the sample.
b) The intrinsic permeability k if the water temperature is 30O C
Q = - K A (h2-H1) / L
K = - Q L / A ∆H = - (1.7) (25) / [(π)(52) (-15) = 0.036 cm/min
Quiz No. 1
CE 331
072
The circular laboratory watershed is 5 meters in diameter. The watershed is subjected to
controlled rainfall resulting in isohyets having the circular shapes as shown. Determine the
average rain over the watershed.
Watershed radius = 5 m
Isohyete Radius = 8 m of 6” rain
10
Isohyete Radius = 4 m of 2” rain
Point Isohyete = 0.4 “ rain
Solution:
Area of watershed= (π)(5)2 =78.54 m2
Isohyete representing the outside area = 2.5”
Isohyete representing inside area = (2+0.4)/2=1.2
Inside area = (π)(4)2 = 50.265 m2
Outside area = 78.54-50.265=28.275 m2
Outside area
Inside area
Precipitation
2.5
1.2
Area
28.275
50.265
Σ78.54
PPT x Area
70.686
60.318
Σ131.004
Average PPt = 131.004/78.54
Quiz No. 2
CE 331
072
A lake received 0.2 inches of rainfall after 24 hours of the storm. Estimate the evaporation
from the lake in inches if the amount of water required to refill a nearby class A pan after the
same period is 0.9 inches.
Solution:
Total Evaporation = (0.2+0.9) (0.7) = 0.77 in
11
Quiz No. 3
CE 331
072
Assume line AB to represent the potential infiltration capacity curve for a given watershed.
Determine the excess rain at the end the second hour of the storm period shown.
Simplified Horton Curve f = 10 - 2.5 t
12 in/hr
A
10 in/hr
8.9
7.5
6.4
4
in/hr
Solution:
0
1
5
2
3
B
4 Time (hr)
F = ∫0-1 (10 - 2.5 t) dt = 10 t – 1.25 t2
4 = 10 t – 1.25 t2 or: t2 – 8t + 3.2 Gives: 0.42 hr
f @ t = 0.42 = 8.9
Runoff = (12) (1hr) – [(12-8.9)+(12-6.4))/2] (1) = 4.3 in
EXAM NO. 1
PROBLEM 1: (10 points)
Compute the total 5 days lake evaporation if the amount of water required to bring the level
from Class A pan to the fixed point are as follows:
____________________________________________________________
Day
1
2
3
4
5
Rainfall (in)
0
0.60
012
0
0.01
Water added
0.30
0.55
0.07
0.28
0.10
Solution:
12
Σ
0.3
1.15
0.19
0.28
0.11 = 2.03 ”
Total lake ppt = (0.7) (2.03) = 1.421 “
PROBLEM 2: (20 points)
Consider a reservoir with one inlet stream, one outlet at a dam and a surface area of .5 km 2.
The reservoir level after weeks of rain is rising at a rate of 3 mm/day. The average
evaporation rate from the reservoir surface is 1.2 mm/day, the inlet discharge is 10,000
m3/day, and the outlet discharge is 16,000 m3/day. Assuming the only variables in the
budget equation are: I,G,O,E (inflow, Groundwater, outflow, evaporation and rate change of
storage dV/dt where I + G – O –E = dV/dt. Find the total net rate of groundwater discharge
into the reservoir
Solution:
dV/dt = + 3 mm/d = 0.003 m/d
Evaporation = 0.0012 m/d
Net discharge = - 6000 m3/d = - [(6000)/(0.5)(1000)2] = 0.012 m/d
Net rate of GW discharge =dv/dt +E+ (O-I) = 0.003 + 0.0012 + 0.012 = 0.016 m/d
PROBLEM 3: (30 points)
Assume line AB to represent the potential infiltration capacity curve for a given watershed.
Determine the excess rain at the end the third hour of the storm period shown.
Simplified Horton Curve f = 10 - 2.5 t
12 in/hr
13
A
10 in/hr
8.95
7.5
C 6.45
4
( in/hr)
D 3.95
2 .5
(In/hr)
0
1
2
3
2.5
B
4 Time (hr)
Solution:
F = ∫0-1 (10 - 2.5 t) dt = 10 t – 1.25 t2
4 = 10 t – 1.25 t2 or: t2 – 8t + 3.2 Gives: 0.42 hr
f @ t = 0.42 = 8.95
Point C = 6.45
Point D = 3.95
Total Runoff = { [(18-8.95) + (18-6.45)]/2} (1 hr) – {[(6.45-2.5)+(3.95-2.5)/2} (1 hr) = 7.6 in
PROBLEM 4: (30 points)
Given the following straight line approximation of infiltration capacity curve and a rain
pattern lasting 4 hours. Determine:
a) The total runoff in cm
b) If the total volume of runoff produced by the storm is 110,000 m3 what is the area of
the watershed in km2?
14
0.2 cm/hr
0.175 cm/hr
A
C
B
0.1 cm/hr
D
0.05 cm/hr
0
1
2
3
(time hr)
Solution:
Total Runoff = A – B + C + D = 0.05 – 0.025 + 0.125 + 0.075 = 0.225 cm or 0.00225 m
Area = 110000/0.00225 = 49,000,000 m2 = 49 km2
EXAM No.2
Problem 1: (60 Points)
Figure below shows a hydrograph from 2 – hr storm. The area of the watershed is 1.0 mi2.
a) Construct a unit hydrograph using straight line separation technique.
15
b) Find the ordinates of 4-hr unit hydrograph.
c) Find the ordinates of a 4-hr hydrograph representing the rain pattern shown in the right.
1 mile = 5280 ft
2 hr effective rain
NO SCALE!
400
Intensity: 0.8”/ hr
NO SCALE!
300
Q
cfs
200
100
0
1
2
3
4
5
6
7
8
9
t ( hr)
4
D.R.O = 1.3”
Q
t
0
1
2
3
4
5
6
7
8
9
10
50
170
300
250
200
150
100
50
0
BF
Q-BF
2-hr
UH
50
50
50
50
50
50
50
50
0
120
250
200
150
100
50
0
0
92.3
192.3
153.8
115.4
76.9
38.5
0
LAG
Σ
0
92.3
192.3
153.8
115.4
76.9
38.5
0
0
92.3
192.3
246.1
307.7
230.7
153.9
76.9
38.5
0
4-hr
UH
X1/2
0
46.15
96.15
123.1
153.8
115.35
76.95
38.45
19.25
0
time (hr)
4-hr
hydr=
X0.8
0
36.92
76.92
98.4
123
92.28
61.56
30.76
15.4
0
Prob. 2: (30 Points)
The actual discharge rates for a flood hydrograph is shown below. The area of the drainage
basin is 3.10 mi2. The rainstorm started at 9 A.M. and ended at 11 A.M.. The base flow is constant at
100 cfs. The volume of direct runoff is 3,600,000 ft3.
Time:
Q:
8 AM
100
9
10 11
100 300 600
12 1 PM
400 200
2
3
100 100
16
a)
b)
c)
d)
At what time did the direct surface runoff begin?
Determine the net rain in inches.
What is the time of concentration of the basin?
What is the time base of hydrograph?
600
500
400
300
200
100
0
8
8 AM
9
10
11
12
1 PM
2
3
4
a)
b)
c)
d)
9
10
100
100
100
100
100
100
11
12
1 PM
0
0
200
500
300
100
0
2
3
RO Started 9 AM
Peak
RO Ends
DRO Began = 9 AM
Net Rain= ½ [ 200+500+300+100] (3600)/(3.1 mi2 x 52802)[12)=0.275 inches
Time of concentration = 3 hrs
Time base = 2+3=5
17