Time (s) Head (ft) 0 2.67 0.1 2.38 0.2 2.13 0.4 1.83 0.8 1.34 1.2 0.65 1.6 -0.02
ID: 2075417 • Letter: T
Question
Time (s)
Head (ft)
0
2.67
0.1
2.38
0.2
2.13
0.4
1.83
0.8
1.34
1.2
0.65
1.6
-0.02
2
-0.55
2.4
-0.96
2.8
-1.26
3.2
-1.45
3.6
-1.47
4
-1.36
4.4
-1.14
4.8
-0.84
5.2
-0.49
5.6
-0.12
6
0.23
6.2
0.53
6.6
0.75
8
0.89
10
0.33
12
-0.6
14
-0.28
16
0.42
18
0.22
20
-0.3
22
-0.2
24
0.21
26
0.16
28
-0.15
30
-0.14
32
0.1
34
0.1
36
-0.08
38
-0.09
40
0.05
42
0.07
44
-0.04
46
-0.05
48
0.02
60
0.01
A slug test was performed on a monitoring well with a radius of 2 in. and a sand pack radius of 5 in. The aquifer thickness was 8 ft. and the initial height of the water column in the casing above the top of the screen was 51 ft. The following data showing the change in the elevation of the water in the casing with time were collected following the lowering of a solid slug into the water. Find the aquifer transmissivity if you assume a storativity of 0.001.
Time (s)
Head (ft)
0
2.67
0.1
2.38
0.2
2.13
0.4
1.83
0.8
1.34
1.2
0.65
1.6
-0.02
2
-0.55
2.4
-0.96
2.8
-1.26
3.2
-1.45
3.6
-1.47
4
-1.36
4.4
-1.14
4.8
-0.84
5.2
-0.49
5.6
-0.12
6
0.23
6.2
0.53
6.6
0.75
8
0.89
10
0.33
12
-0.6
14
-0.28
16
0.42
18
0.22
20
-0.3
22
-0.2
24
0.21
26
0.16
28
-0.15
30
-0.14
32
0.1
34
0.1
36
-0.08
38
-0.09
40
0.05
42
0.07
44
-0.04
46
-0.05
48
0.02
60
0.01
Explanation / Answer
given:
aquifer thickness b=8ft
initial displacement H0=51ft
storativity S=.001
the aquifer trasmittivity T=kb
where k=hydraulic conductivity
we get k from the graph of normalized head vs time where normalized head =H/Ho
as an example from given data at t=0s the normalized head =2.67/51=.052
now we draw the graph to get t0 for calculation of k
k=rc^2ln(Le/rw)/2.Le.t0
now rc=5in, Le=length of well=8ft, rw=2in, t0=time lag from graph
then k=5^2ln(8/5)/2.8.t0=.734/t0
using this in T=k.b=.734/to*8 is the transmittivity where we get t0 from graph.
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