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Homework 4 Groundwater Hydrology

1. a. Confined Aquifer:

𝑟! = 3.3  𝑚;    𝑟! = 11  𝑚;    𝑄 = 1250   !!"#

;    ℎ! −  ℎ! =    22− 15  𝑚 = 7  𝑚

𝑇 =  𝑄

2𝜋(ℎ! −  ℎ!)ln

𝑟!𝑟!

=  1250   𝑙

𝑚𝑖𝑛. 001  𝑚!

𝑙2𝜋(7𝑚) ln

113 = .0342  

𝑚!

𝑚𝑖𝑛

= 𝐾𝑏 = 𝐾(10  𝑚)

 𝐾 =. 034210

𝑚𝑚𝑖𝑛 = 3.42 ∗  10!!  

𝑚𝑚𝑖𝑛

b. Unconfined Aquifer – Thiem Solution with:

𝑟! = 2.2  𝑚;    𝑟! = 7.9  𝑚;    𝑄 = 750  𝑙

𝑚𝑖𝑛 ;    

𝑏!! −  𝑏!! =   ((10− 1)! −  (10− 4)!)𝑚! =  45  𝑚!

𝐾 =  𝑄

𝜋(𝑏!! −  𝑏!!)ln

𝑟!𝑟!

=  750   𝑙

𝑚𝑖𝑛. 001  𝑚!

𝑙𝜋(45  𝑚!) ln

7.92.2 =  6.78 ∗  10!!  

𝑚𝑚𝑖𝑛

c. If you look carefully at the transient solution steady state will never be attained.

However at some point the rate of change of head with respect to time will be so small that your measurement device will no longer notice changes.

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2. Analytical approximation (Graphical Method provides equivalent results, but not shown here – see example in class notes for method):

Here is the data:

For well 1 it looks like ∆𝑠!" = 5.8− 2.8 = 3  𝑚 and 𝑡! =   .14  𝑚𝑖𝑛, so

𝑇 =   .183320   𝑚

!

𝑚𝑖𝑛3  𝑚 = 1.22  

𝑚!

𝑚𝑖𝑛

𝑆 = 2.2461.22   𝑚

!

𝑚𝑖𝑛 ∗ .14  𝑚𝑖𝑛2!  𝑚! =   .096

For well 2 it looks like ∆𝑠!" = 4.9− 2.5 = 2.4  𝑚 and 𝑡! =  1  𝑚𝑖𝑛, so

𝑇 =   .183320   𝑚

!

𝑚𝑖𝑛2.4  𝑚 = 1.53  

𝑚!

𝑚𝑖𝑛

𝑆 = 2.2461.53   𝑚

!

𝑚𝑖𝑛 ∗ 1  𝑚𝑖𝑛5!  𝑚! =   .137

ii. Depending on your perspective one could see these measurements as consistent or not consistent. It depends what you find acceptable as error.

10-1 100 101 1020

1

2

3

4

5

6

7

8

9

10

Time t

Draw

down

s

Well 1Well 2

3

If one deems them inconsistent it could because one of the many assumption associated with this analytical solution could be violated. For example if the aquifer were heterogeneous something like this might happen.

3. i. I would us the overdamped Hvorslev method, since the data from all three tests

is monotonically decreasing and looks exponential. Also, given that we are told later that the well is not fully penetrating it seems the best approach.

ii. Test 1: 79.88008996 ∗ 37% =  29.55, which for all data sets occurs somewhere between t = 50 s and t = 90 s. To choose which value to use between 50 and 100, I used this interpolation formula on the actual data:

𝑡!" =  𝐻!" −  𝐻!"#$%𝐻!"#$% −  𝐻!"#$%

𝑡!"#$% − 𝑡!"#$% + 𝑡!"#$%

where H37 is 29.55, Hbelow is the first head measurement in the data which is first less than H37, and Habove is the preceding on. tabove and tbelow are the time associated with theses respectively You may also just read it off a graph that you generate Approximately For test 1, 𝑡!" =  71. For test 2, 𝑡!" =  59. For test 3, 𝑡!" =  82.

Now we can solve for the hydraulic conductivity:

𝐾   =𝑟! ln 𝐿!

𝑅2𝐿!𝑡!"

With 𝑅 =   .5  𝑚; 𝑟 =   .6  𝑚;𝑎𝑛𝑑  𝐿! = 4  𝑚.

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Test 1:

𝐾   =(.6  𝑚)! ln 4  𝑚

. 5  𝑚2 4  𝑚 (71  𝑠) =  0.0013  

𝑚𝑠 =  1.3  

𝑚𝑚𝑠

Test 2:

                 𝐾   =(.6  𝑚)! ln 4  𝑚

. 5  𝑚2 4  𝑚 (59  𝑠) =  0.0016  

𝑚𝑠 = 1.6  

𝑚𝑚𝑠

Test 3:

         𝐾   =(.6  𝑚)! ln 4  𝑚

. 5  𝑚2 4  𝑚 (82  𝑠) =  0.0011  

𝑚𝑠 = 1.1  

𝑚𝑚𝑠

So the slug test, using the overdamped Hvorslev method gives us a hydraulic conductivity estimate of:

1.1  𝑚𝑚𝑠  ≤ 𝐾   ≤ 1.6  

𝑚𝑚𝑠 .

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4. Here is the overlaid data:

This suggests that mu matches closest with the curve for mu=1e-2 It also appears that t1 is at about 40s.

Now we can solve:

𝑇 =  𝑟!!

𝑡!=  (2.54  𝑐𝑚)!

40 =   .16  𝑐𝑚!

𝑠  

𝑆 =  𝑟!!𝜇𝑟!!

=  (2.54  𝑐𝑚)! ∗ 10!!

(2.54  𝑐𝑚)! =  10!!

Note that if you answer does not match this exactly it is ok – this method is extremely sensitive to error.

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5. i. We need to use the underdamped Van der Kamp method, because the head oscillates.

ii. Here is the data:

We can estimate

𝜔 =  2𝜋

(19− 6)  𝑠 =2𝜋13  𝑠 =   .4833  𝑠

!!

𝛾 =  ln −50

−30(19− 6)𝑠 =  

ln 53

13  𝑠 =   .0393  𝑠!!

𝐿 =  𝑔

𝜔! +  𝛾! =  9.81  𝑚𝑠!

2𝜋13  𝑠

!+  

𝑙𝑛 53

13  𝑠

! =  41.72  𝑚

𝑑 =  𝛾𝑔/𝐿

=  . 4833/𝑠9.81

41.72  𝑠!

=   .0810  

𝑎 =  𝑟!! 𝑔/𝐿8𝑑 =  

(2.54  𝑐𝑚)! 9.8141.72𝑠!

8 ∗ .0810 = 4.826  𝑐𝑚!

𝑠  

0 5 10 15 20 25 30-60

-40

-20

0

20

40

60

t (s)

h (c

m)

Question 5 data

data spline

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𝑐 =  −𝑎𝑙𝑛 . 79𝑟!!𝑆𝑔𝐿 =  −4.826 ∗ ln . 79 ∗ 2.54! ∗ 2 ∗ 10!!

9.8141.72

𝑐𝑚!

𝑠 =  47.849  𝑐𝑚!

𝑠

Now we can plug 𝑐 into our algorithm to find T (below is a Matlabl script that does this for you – but doing it by hand does the same thing).

c = 47.849; a = 4.826; temp = c+a*log(c); while temp > c+a*log(temp) + 10^-4 || temp < c+a*log(temp) - 10^-4 temp = c+a*log(temp); end temp

We get T = 68.2285 !"!

!.