Homework 3
2006 Hydrogeology
Due 5/18/2006
1.
(a) A community is installing a new well in a regionally confined aquifer with
a transmissivity of 143 m2/day and a storativity of 0.0005. The planned
pumping rate is 0.0196 m3/sec. There are several nearby wells tapping the
same aquifer, and the project manager needs to know if the new well will cause
significant interference with these wells. Compute the theoretical drawdown
caused by the new well after 30 days of continuous pumping at the following
distances: 15, 45, 75, 150, 300, 900, 1800, and 3000 m.
(b) Use EXCEL to plot the distance-drawdown data from (a) on semilog paper.
(c) With reference to the well and aquifer system in (a), compute the drawdown at a
distance of 75 m at the following times: 1, 2, 5, 10, 15, 30, and 60 min; 2, 5, and
12 h; and 1, 5, 10, 20, and 30 days.
(d) Use EXCEL to plot the time-drawdown data from (c) on semilog paper.
2.
A well that pumps at a constant rate of 2106 m3/day has achieved equilibrium so that
there is no change in the drawdown with time. (The cone of depression has expanded
to include a recharge zone equal to the amount of water being pumped.) The well taps
a confined aquifer that is 5.4 m thick. An observation well 37.5 m away has a head of
83 m above sea level; another observation well 116 m away has a head of 87 m.
Compute the value of aquifer transmissivity using the Thiem equation.
3.
The following data are from a pumping test where a well was pumped at a rate of
0.276 m3 per minute. Drawdown as shown was measured in an observation well
75 m away from the pumped well. The geologist’s log of the well is as follows:
0 ~ 6.9 m
6.9 ~ 23.1 m
23.1 ~ 54.6 m
54.6 ~ 65.1 m
65.1 ~ 66.3 m
Glacial till, brown, clayey
Dolomite, fractured
Shale, black, dense
Sandstone, well cemented, coarse
Shale, gray, limy
Time
Drawdown
Time
Drawdown
Time
Drawdown
Time
Drawdown
(min)
(m)
(min)
(m)
(min)
(m)
(min)
(m)
0
0.000
5
0.447
24
0.708
120
0.984
1
1.5
2.0
2.5
3.0
4.0
0.198
0.261
0.297
0.333
0.363
0.408
6
8
10
12
14
18
0.477
0.525
0.558
0.591
0.624
0.66
30
40
50
60
80
100
0.747
0.795
0.834
0.864
0.912
0.948
150
180
210
240
1.026
1.053
1.083
1.101
-1-
Homework 3
2006 Hydrogeology
Due 5/18/2006
A steel well casing was cemented to a depth of 54.6 m and the well was extended
as an open boring past that point.
(a) Plot the time-drawdown data on 3× 5 cycle logarithmic paper. Use the Theis
type curve to find the aquifer transmissivity and storativity. Compute the
average hydraulic conductivity.
(b) Replot the data on four-cycle semilogarithmic paper. Use the Cooper-Jacob
straight-line method to find the aquifer transmissivity and storativity.
(Use the Theis type curve in page 3 and the blank logarithmic and
semilogarithmic papers in pages 4 and 5)
4.
A slug test was made with a piezometer that had a casing radius of 2.54 cm and a
screen of radius 2.54 cm. A slug of 4000 cm3 of water was injected; this raised
the water level by 197.3 cm. The well completely penetrated a confined stratum
that was 2.3 m thick. The decline in head with time is given in the following
chart.
Time (sec)
Drawdown (cm)
Time (sec)
Drawdown (cm)
0
197.3
22
118.4
1
2
3
5
7
10
13
17
185.4
178.6
173.6
167.7
158.8
147.0
140.0
129.2
32
53
84
119
170
245
400
800
99.6
74.0
51.3
35.5
23.3
15.2
8.7
4.3
Plot the data on semilogarithmic paper and find the aquifer transmissivity and
storativity using the Cooper-Bredehoeft-Papadopulos method (Use the
semilogarithmic paper in page 6, and see the Cooper-Bredehoeft-Papadopulos
type curve in page 7)
-2-
Homework 3
2006 Hydrogeology
Due 5/18/2006
101
W(u)
100
10-1
10-2
10-3 -1
10
100
101
102
103
104
1/u
-3-
Homework 3
2006 Hydrogeology
Due 5/18/2006
Drawdown (m)
101
10
0
10-1
10-2 -1
10
10
0
1
10
10
Time (min)
2
10
3
10
4
-4-
Homework 3
2006 Hydrogeology
Due 5/18/2006
2
Drawdown (m)
1.5
1
0.5
0 -1
10
100
101
Time (min)
102
103
-5-
Homework 3
2006 Hydrogeology
Due 5/18/2006
1
0.8
H/H0
0.6
0.4
0.2
0 -1
10
100
101
102
Time (sec)
103
104
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Homework 3
2006 Hydrogeology
Due 5/18/2006
-7-
Homework 3
2006 Hydrogeology
Due 5/18/2006
-8-
Homework 3
2006 Hydrogeology
Due 5/18/2006
Well function data
1/u
W(u)
1/u
W(u)
1/u
W(u)
1/u
W(u)
1/u
W(u)
1/u
W(u)
1.00E-01
1.50E-01
2.00E-01
2.50E-01
3.00E-01
3.50E-01
4.00E-01
0
0
0.001
0.004
0.009
0.016
0.025
1.00E+00
1.50E+00
2.00E+00
2.50E+00
3.00E+00
3.50E+00
4.00E+00
0.219
0.398
0.56
0.702
0.829
0.942
1.044
1.00E+01
1.50E+01
2.00E+01
2.50E+01
3.00E+01
3.50E+01
4.00E+01
1.823
2.196
2.468
2.681
2.857
3.007
3.137
1.00E+02
1.50E+02
2.00E+02
2.50E+02
3.00E+02
3.50E+02
4.00E+02
4.038
4.44
4.726
4.948
5.13
5.284
5.417
1.00E+03
1.50E+03
2.00E+03
2.50E+03
3.00E+03
3.50E+03
4.00E+03
6.332
6.737
7.024
7.247
7.429
7.584
7.717
1.00E+04
1.50E+04
2.00E+04
2.50E+04
3.00E+04
3.50E+04
4.00E+04
8.633
9.039
9.326
9.549
9.732
9.886
10.019
4.50E-01
5.00E-01
5.50E-01
6.00E-01
6.50E-01
7.00E-01
7.50E-01
8.00E-01
8.50E-01
9.00E-01
0.036
0.049
0.063
0.078
0.094
0.111
0.129
0.146
0.164
0.183
4.50E+00
5.00E+00
5.50E+00
6.00E+00
6.50E+00
7.00E+00
7.50E+00
8.00E+00
8.50E+00
9.00E+00
1.137
1.223
1.301
1.375
1.443
1.507
1.567
1.623
1.677
1.728
4.50E+01
5.00E+01
5.50E+01
6.00E+01
6.50E+01
7.00E+01
7.50E+01
8.00E+01
8.50E+01
9.00E+01
3.252
3.355
3.448
3.534
3.612
3.686
3.754
3.817
3.877
3.934
4.50E+02
5.00E+02
5.50E+02
6.00E+02
6.50E+02
7.00E+02
7.50E+02
8.00E+02
8.50E+02
9.00E+02
5.534
5.639
5.735
5.821
5.901
5.975
6.044
6.109
6.169
6.226
4.50E+03
5.00E+03
5.50E+03
6.00E+03
6.50E+03
7.00E+03
7.50E+03
8.00E+03
8.50E+03
9.00E+03
7.835
7.94
8.035
8.122
8.202
8.277
8.346
8.41
8.471
8.528
4.50E+04
5.00E+04
5.50E+04
6.00E+04
6.50E+04
7.00E+04
7.50E+04
8.00E+04
8.50E+04
9.00E+04
10.137
10.243
10.338
10.425
10.505
10.579
10.648
10.713
10.773
10.83
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