Homework 1 Solution
1.
2006 Hydrogeology
Due 3/14/2006
At a water elevation of 6391 ft, Mono Lake has a volume of 2,939,000 ac-ft, and
a surface area of 48,100 ac. Annual inputs to the lake include 8 in. of direct
precipitation, runoff from gauged streams of 150,000 ac-ft per year, and
ungauged runoff and groundwater inflow of 37,000 ac-ft per year. Evaporation is
45 in. per year.
(A) Make a water budget showing inputs, in ac-ft per year and outputs in ac-ft
per year. Does the input balance the output?
(B) Will the average lake level rise or fall from the 6391-ft elevation over the
long term?
(C) What would be the lake surface area when the inputs balance the
outputs?(Assume that the volume of gauged and ungauged runoff and
ground-water inflows remain constant with a change in lake surface area. )
(D) What is the residence time* for water in Mono Lake when the water surface
is at 6391 ft?
*The residence time of a body of water is the average time that is would take for
the volume of water to be exchanged once.
Solution:
First convert all measurements to units of acre-feet per year
Precipitation = 8 in/yr × 0.083 ft/in = 0.67 ft/yr
= 0.67 ft/yr × 48,100 ac = 32,067 ac-ft/yr
Evaporation = 45 in/yr × 0.083 ft/in = 3.75 ft/yr
= 3.75 ft/yr × 48,100 ac = 180,375 ac-ft/yr
(A)
INPUTS(ac-ft/yr)
Precipitation
32,000
Streamflow
150,000
Ground water inflow
Total
37,000
219,000
OUTFLOWS (ac-ft/yr)
Evaporation
180,000
Total
180,000
Input is greater than output
(B)
The lake level will rise over the long term, as inputs are greater than the only output,
which is evaporation.
(C)
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Homework 1 Solution
2006 Hydrogeology
Due 3/14/2006
In order for the inputs and outputs to be equal, the lake will rise so that the surface
area is greater and more water evaporates as evaporation as well as the precipitation
input is a function of lake surface area.
The excess input at an area of 48,100 acre surface area is 219,000 ac-ft/yr – 180,000
ac-ft/yr or 39,000 ac-ft/yr
The net evaporation is total evaporation less precipitation
Net evaporation = 45 in/yr – 8 in/yr = 37 in/yr = 3.08 ft/yr
Increase in area = (39,000 ac-ft/yr) / (3.08 ft/yr) = 12,649 ac
Total area of lake = 12, 662 ac + 48,100 ac = 60,762 ac = 61,000 ac
(D)
The residence time = (2,939,000 ac-ft) / (180,000 ac-ft/yr) = 16.3 yr
2.
Figure 2.28 is the hydrograph of a river with a long summer baseflow recession.
Compute the volume of annual recharge that occurs between runoff year 1 and
runoff year 2.
Solution:
Qo in year 1 is 280 ft3/s
0.1 Qo in
year 1 is reached in 5.5 months
-2-
Homework 1 Solution
2006 Hydrogeology
Due 3/14/2006
t1 = 5.5 mo× 8.64× 104 sec/day× 30 day/mo = 1.4256× 107 s
Vtp = Qo t1 / 2.3026
Vtp = (280× 1.4256× 107) / 2.3026 = 1.73 × 109 ft3
Total recession in year 1, t, is 6.4 months.
Vt = Vtp / 10t(t/t1)
Vt = 1.73× 109 ft3 / 10(6.4/5/5)
Vt = 1.19× 108 ft3 Where Vt is the volume of water that could contribute to
baseflow at the end of the first year’s recession.
Qo in year 2 is 320 ft3/s
0.1 Qo is reached in 5.9 mo
t2 is 5.9 mo or 1.53× 107 s
Vtp2 = (320 ft3/s× 1.53× 107s) / 2.3026 = 2.13× 109 ft3
The amount of recharge between recessions is Vtp2 -Vt
Vtp2 -Vt = 2.13× 109 ft3 - 1.19× 108 ft3
Vtp2 -Vt = 2.00× 109 ft3
3.
A constant-head permeameter has a cross-sectional area of 78.5 cm². The sample
is 23 cm long. At a head of 3.4 cm, the permeameter discharges 50 cm³ in 38s.
(A) What is the hydraulic conductivity in centimeters per second and feet peer
day?
(B) What is the intrinsic permeability if the hydraulic conductivity was measured
at 15℃?
(C) From the hydraulic conductivity value, name the type of soil.
Solution:
(A)
K=
VL
Ath
Where V = 50 cm 3
L = 23 cm
A = 78.5 cm 2
t = 38 s
h = 3.4 cm
-3-
Homework 1 Solution
K=
2006 Hydrogeology
Due 3/14/2006
50cm3 i23cm
= 0.11 cm s = 316.8 ft d
78.5cm 2 i38s i3.4cm
(B)
At 15 Ｃ,ρ =0.999099 g / cm3 and µ = 0.011404 g / (cm ⋅ s )
k=
0.11cm / s × 0.011404 g /(cm ⋅ s )
Kµ
⇒k=
= 1.3×10−6 cm 2
3
2
ρg
0.999099 g / cm ×980cm / s
(C)
K = 0.11cm / s, soil is well sorted sand or gravel.
An aquifer has a specific yield of 0.19. During a drought period, the following
average declines in the water table were noted:
Area
Size (km²)
Decline (m)
A
15
2.34
B
7.5
1.22
C
18.3
0.76
D
22.5
3.44
E
9.44
1.89
F
22.7
0.35
What was the total volume of water represented by the decline in the water table?
Solution:
4.
Vw = SA∆h
Where S = 0.19
AA = 15 km 2 = 1.5 × 10 7 m 2
AB = 7.5 km 2 = 7.5 × 10 6 m 2
AC = 18.3 km 2 = 1.83 × 10 7 m 2
AD = 22.5 km 2 = 2.25 × 10 7 m 2
AE = 9.44 km 2 = 9.44 × 10 6 m 2
AF = 22.7 km 2 = 2.27 × 10 7 m 2
and
∆h A = 2.34 m
∆hB = 1.22 m
∆hC = 0.76 m
∆hD = 3.44 m
∆hE = 1.89 m
∆hF = 0.35 m
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Homework 1 Solution
2006 Hydrogeology
Due 3/14/2006
Area A :VWA = 0.19 ⋅ 1.5 × 10 7 m 2 ⋅ 2.34 m
VWA = 6.67 × 10 6 m 3
VWB = 1.74 × 10 6 m 3
VWC = 2.64 × 10 6 m 3
VWD = 1.47 × 10 7 m 3
VWE = 3.39 × 10 6 m 3
VWF = 1.51 × 10 6 m 3
Total volume of water the decline of head in the aquifer is 3.07×107 m3
5. A confined aquifer has a specific storage of 7.239×10−3 m−1 and a porosity of
0.274. The compressibility of water is 4.6×10−10 m 2 / N . What is the
compressibility of the aquifer skeleton?
Solution:
S s = ρ g (α + n β ) = 7.239 ×10−3 m−1
= 1000 kg / m 3 × 9.81 m / s 2 ×(α + 0.274 × 4.6 ×10−10 m 2 / N )
⇒ α = 7.380 ×10−7 m 2 / N
6.
An aquifer has three different formations. Formation A has a thickness of 22 ft
and a hydraulic conductivity of 17.0 ft/d. Formation B has a thickness of 3.5 ft
and a conductivity of 99 ft/d. Formation C has a thickness of 26 ft and a
conductivity of 22 ft/d. Assume that each formation is isotropic and
homogeneous. Compute both the overall horizontal and vertical conductivities.
Solution:
Fm
K ( ft / d )
bm ( ft )
A
B
17.0
99
22
3.5
C
22
26
and total thickness = 51.5 ft
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Homework 1 Solution
n
K haverage =
∑
2006 Hydrogeology
Due 3/14/2006
khm bm
m=1
=
K vaverage =
b
17.0 ft d i22 ft 99 ft d i3.5 ft 22 ft d i26 ft
+
+
= 25 ft d
51.5 ft
51.5 ft
51.5 ft
b
n
∑ Kb
m
m=1
=
7.
vm
51.5 ft
= 21 ft d
22 ft
3.5 ft
26 ft
+
+
17.0 ft d 99 ft d 22 ft d
Use the Hazen method to estimate the hydraulic conductivity of the sediments
graphed in Figure 3.33.
Solution:
K = C (d10 ) 2
d10 = 0.16mm×
1cm
= 0.016cm
10mm
C 100
K = 100× (0.016) 2 = 2.6×10−2 cm / s
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Homework 1 Solution
8.
2006 Hydrogeology
Due 3/14/2006
Given the following set of data representing the hydraulic conductivity of core
samples from the same formation, perform the following:
Sample #
K (m/s)
4.3×10−4
1
2
6.1×10−3
3
2.5×10−5
4
1.2×10−4
5
1.0×10−6
6
7.1×10−3
7
9.1×10−6
8
2.2×10−3
9
4.2×10−5
10
8.7×10−4
11
3.5×10−5
(A) Find the arithmetic mean of the data set.
(B) Find the geometric mean of the data set.
(C) Make a histogram of the data set.
(D) Make a histogram of the log transformed data.
-7-
Homework 1 Solution
Solution:
Sample #
2006 Hydrogeology
Due 3/14/2006
K (m/s)
ln (K)
K bin
ln (K) bin
1
4.30E-04
1.00E-03
−4
2
6.10E-03
− 7.75
− 5.10
5.00E-04
−5
3
2.50E-05
1.00E-04
4
1.20E-04
5
1.00E-06
− 10.60
− 9.03
− 13.82
−6
−7
−8
6
7.10E-03
5.00E-06
7
9.10E-06
8
2.20E-03
− 4.95
− 11.61
− 6.12
9
4.20E-05
10
8.70E-04
11
3.50E-05
− 10.08
− 7.05
− 10.26
1.69E-02
− 96.35
sum
5.00E-05
1.00E-05
−9
− 10
1.00E-06
− 11
− 12
Arithmetic mean = 1.54E - 03m s
Geometric mean = 1.57E - 04 m s
histogram data for K
histogram data for ln (K)
Bin
Frequency
Bin
Frequency
1.E-06
1
− 12
1
5.E-06
0
− 11
1
1.E-05
1
3
5.E-05
3
1.E-04
0
− 10
−9
−8
5.E-04
2
2
1.E-03
1
> 5.E - 3
3
−7
−6
−5
−4
1
1
0
1
1
-8-
Frequency
Homework 1 Solution
2006 Hydrogeology
Due 3/14/2006
3.5
3
2.5
2
1.5
1
0.5
0
1.00E-06 5.00E-06 1.00E-05 5.00E-05 1.00E-04 5.00E-04 1.00E-03 >0.005
Frequency
K (cm/s)
3.5
3
2.5
2
1.5
1
0.5
0
-12
-11
-10
-9
-8
-7
-6
-5
-4
ln(K)
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