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Homework 4 Groundwater Hydrology
1. a. Confined Aquifer:
π! = 3.3 π; π! = 11 π; π = 1250 !!"#
; β! β β! = 22β 15 π = 7 π
π = π
2π(β! β β!)ln
π!π!
= 1250 π
πππ. 001 π!
π2π(7π) ln
113 = .0342
π!
πππ
= πΎπ = πΎ(10 π)
πΎ =. 034210
ππππ = 3.42 β 10!!
ππππ
b. Unconfined Aquifer β Thiem Solution with:
π! = 2.2 π; π! = 7.9 π; π = 750 π
πππ ;
π!! β π!! = ((10β 1)! β (10β 4)!)π! = 45 π!
πΎ = π
π(π!! β π!!)ln
π!π!
= 750 π
πππ. 001 π!
ππ(45 π!) ln
7.92.2 = 6.78 β 10!!
ππππ
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.
2
2. Analytical approximation (Graphical Method provides equivalent results, but not shown here β see example in class notes for method):
Here is the data:
For well 1 it looks like βπ !" = 5.8β 2.8 = 3 π and π‘! = .14 πππ, so
π = .183320 π
!
πππ3 π = 1.22
π!
πππ
π = 2.2461.22 π
!
πππ β .14 πππ2! π! = .096
For well 2 it looks like βπ !" = 4.9β 2.5 = 2.4 π and π‘! = 1 πππ, so
π = .183320 π
!
πππ2.4 π = 1.53
π!
πππ
π = 2.2461.53 π
!
πππ β 1 πππ5! π! = .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 β 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:
π‘!" = π»!" β π»!"#$%π»!"#$% β π»!"#$%
π‘!"#$% β π‘!"#$% + π‘!"#$%
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, π‘!" = 71. For test 2, π‘!" = 59. For test 3, π‘!" = 82.
Now we can solve for the hydraulic conductivity:
πΎ =π! ln πΏ!
π 2πΏ!π‘!"
With π = .5 π; π = .6 π;πππ πΏ! = 4 π.
4
Test 1:
πΎ =(.6 π)! ln 4 π
. 5 π2 4 π (71 π ) = 0.0013
ππ = 1.3
πππ
Test 2:
πΎ =(.6 π)! ln 4 π
. 5 π2 4 π (59 π ) = 0.0016
ππ = 1.6
πππ
Test 3:
πΎ =(.6 π)! ln 4 π
. 5 π2 4 π (82 π ) = 0.0011
ππ = 1.1
πππ
So the slug test, using the overdamped Hvorslev method gives us a hydraulic conductivity estimate of:
1.1 πππ β€ πΎ β€ 1.6
πππ .
5
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:
π = π!!
π‘!= (2.54 ππ)!
40 = .16 ππ!
π
π = π!!ππ!!
= (2.54 ππ)! β 10!!
(2.54 ππ)! = 10!!
Note that if you answer does not match this exactly it is ok β this method is extremely sensitive to error.
6
5. i. We need to use the underdamped Van der Kamp method, because the head oscillates.
ii. Here is the data:
We can estimate
π = 2π
(19β 6) π =2π13 π = .4833 π
!!
πΎ = ln β50
β30(19β 6)π =
ln 53
13 π = .0393 π !!
πΏ = π
π! + πΎ! = 9.81 ππ !
2π13 π
!+
ππ 53
13 π
! = 41.72 π
π = πΎπ/πΏ
= . 4833/π 9.81
41.72 π !
= .0810
π = π!! π/πΏ8π =
(2.54 ππ)! 9.8141.72π !
8 β .0810 = 4.826 ππ!
π
0 5 10 15 20 25 30-60
-40
-20
0
20
40
60
t (s)
h (c
m)
Question 5 data
data spline
7
π = βπππ . 79π!!πππΏ = β4.826 β ln . 79 β 2.54! β 2 β 10!!
9.8141.72
ππ!
π = 47.849 ππ!
π
Now we can plug π into our algorithm to find T (below is a Matlabl script that does this for you β 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 !"!
!.
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