Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote