Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
Vadoze Zone Hydrology
Introduction
In this laboratory, we have to determine the saturated hydraulic conductivity (Ks), by
measurement of the constant head method. In the first case, we have a one-layer soil column,
and three series of data.
Then we measure another hydraulic conductivity (Ks) through an hydraulic head distribution
within a two-layer soil column.
We’ve effected the experiment as described in the procedure and in the following figure.
Discuss of saturated hydraulic conductivity (Ksat) and methods of measurement
Ks(θ) = a + eb.θ,
And we have
where a,b are specific constant of the soil.???????
Ksat = Ks(θs) = a + eb.θs.
MANQUE ENCORE UN PEU DE TEXTE SUR KSAT, KéKE C’EST ET COMMENT LE
DETERMINER
1
Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
Plot of the calculated hydraulic conductivity Ks vs. Time for a one-layer soil column
Ks vs. time
0.70
Ks A 12
Ks A 23
0.60
Ks A 13
0.50
Ks [m/s]
Ks B 12
0.40
Ks B 23
0.30
Ks B 13
Ks C 12
0.20
Ks C 23
0.10
Ks C 13
0.00
18
38
62
85
107
127
150
172
196
219
Time [s]
We have calculated the Ks coefficient three times for the one-layer soil column with different
height of the bubbling tube as describe in the tab XX.
[cm]
height H(stock)
diameter
elevation
thickeness
H(sol)
z*(H1)
z(H2)
z(H3)
SAND
CLAY
41.5 17.5 51
51
6
6
6
6
8.5 8.5 8.5 8.5
20
2
10
18
20
2
10
18
20
2
10
18
9.5
2
5
8.2
* outlet piezometer's water, elevated from soil's base
For these three heights, we are able to determine several Ks between each piezometrical
measure’s point with Darcy’s law, as describe here :
Ks = - Jw * Δz / Δψh
where ψh = ψm + ψp + ψz = ψp + ψz .
We observe that the approximate value of Ks 23 is the same for the different ABC case. We
can do the same observation for Ks 13, but the concordance isn’t well as before. In fact, there
is a big difference between the three Ks 12.
2
Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
By looking the previous graph, we remark a general trend : Ks 23 is smaller than Ks 12, so it
result that Ks 13 is situated between both.
Theoretically, we should obtain a same value for all this Ks, because we have
experimentations over an one-layer soil column, but we don’t.
A first global explanation is that there was air in the system for the two first experimentations,
so the continuity pressure wasn’t assured. Consequently, we cannot use the Darcy’s law with
our measurements, which aren’t more representative.
Another explanation could be our inability of good precision during measurement. The water
level is not easy to read exactly at 10 ml in the graduate cylinder, water outflow making move
level.
Secondly, Ks 12 is different than Ks 23 because we think that the piezometers gave us a
wrong value. It’s also possible that some preferential path have been formed in the section 12.
For the next graph, we determined the Ks effective using the following equation:
Keff = Σ Li / Σ ( Li / Ksi ).
We use Ks 12 and Ks 23 in this formula and the result should be equal to Ks 13. This relation
is respected : Ks eff = Ks 13 for the three experiment.
Because we don’t obtain the same result for Ks 12 and Ks 23 alone, we can suppose that the
piezometer 2 give us a wrong value, Ks eff calculated corresponding exactly to Ks 13. It
assumes that Ks 12 and Ks 23 are wrong calculated. However, when we determine Ks eff, this
false measurement at piezometer 2 disappears when we use Darcy’s law.
Once again, there isn’t a good exactitude between this three Ks, for the same reasons
probably, as previously said. In fact, analysing Ks eff of the first experimentation, we see a
big fluctuation of its value vs. time. It’s the sign there was air in the system and the inflow
water pressure fluctuated. Ks eff of the two other heads are more constant during the time, it’s
probably another reason of their non-exactitude.
Ks eff vs. time
0.300
Ks A eff
0.250
Ks B eff
Ks [m/s]
0.200
0.150
Ks C eff
0.100
0.050
0.000
1
2
3
4
5
6
Time [s]
7
8
9
10
3
Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
Plot of the calculated hydraulic conductivity Ks vs. Time for a two-layer soil column
We have calculated the Ks coefficient one time for the two-layer soil column with a single
height of the bubbling tube as describe in the tab XX before. Water took to much time to
circule trough this soil and we haven’t enough time to do other experimentation with this
column. More over, there was air in piezometer 3, so we cannot use its results for our calcul.
We have define Ks 12 as describe here down.
First we determine the flux Jw, using the measurment of soil’s water outflow versus time.
Then using the Darcy’s law between point 1 and 2, we can calculate the Ks 12 coefficient.
As see before, unknown piezometer value in measure point 3 doesn’t let us define Ks 23 and
Ks 13 in the same way.
Because we have some value for the bubbling tube, height of soil column, atmospheric
pressure and water outflow, we are able to calculate the Ks eff coefficient, which should be
equals to the Ks 13 coefficient. Then we can define the Ks 23, using the Ks eff and Ks 1to3
relation.
At the top of the soil column, we suppose the pressure equals to the atmospheric pressure.
A CALUCLER
Plot a potential diagram of the two-layer experiment
Compute the Ksat and Ks for each layer
4
Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
Discuss Ksat vs. If you thought it was a one-layer  Ks-eff
For the 3 first case, this law is respected. But with the two-layer soil column, because there
was air in the system piezometer 2 doesn’t give us valide resultat. cannot calculate Keff .
Probably if we had results for this last case, there would be confirme this relationship between
Keff and Ki .
5
Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
For different situation we obtain the following results :
SABLE
ARGILE
Hstock
[ml]
41.5
17.5
[min.sec / 10 ml]
51
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
0.18
0.38
1.02
1.25
1.47
2.07
2.30
2.52
3.16
3.39
0.16
0.32
0.48
0.64
0.80
0.96
1.12
1.28
1.44
0.51
1.40
2.31
3.21
4.10
5.01
5.52
SAND
51
[min.sec]
[ml]
3.38
8.02
11.05
13.11
15.28
5.5
7.0
8.5
9.5
10.5
CLAY
With this different parameters
[cm]
height H(stock)
diameter
elevation
thickeness H(sol)
z*(H1)
z(H2)
z(H3)
SAND
41.5
6
8.5
20
2
10
18
17.5
6
8.5
20
2
10
18
CLAY
51 51
6
6
8.5 8.5
20 9.5
2
2
10 5
18 8.2
* outlet piezometer's water, elevated from soil's base
We can measure and calculate
Piezometer height
H1
H2
H3
43
36
6.5
36
31
10
66.5
48
11.5
70.5
70.5
AIR
Piezometer value Pi = Hi - z(Hi)
P1
P2
P3
41
26
-11.5
34
21
-8
64.5
38
-6.5
68.5
65.5
-
6
Berner Nicolas
Brethaut Yoann
Gafsou Blaise
Lochmatter Samuel
LABORATORY 2
december, 12 2006
41.5
12.00
17.5
51
eau écoulée [ml]
10.00
51
8.00
6.00
4.00
2.00
0.00
0.0
20.0
40.0
60.0
80.0
100.0
temps [min]
First we have had a problem, there was air in the tube of arrival water. The pressure was’nt at
saturation.
7