# 105328899-7-Hydraulic-Conductivity

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```					Methods Used to Determine
Hydraulic Conductivity /
Permeability
Background
• Hydraulic Conductivity, K, is essential to
understanding flow through soils.
– Darcy’s Law
– Richards’ Equation
• Soil characteristics that determine K
– Particle size
– Porosity
– Bulk density
• K is a function of pressure or moisture
content
– low matric potential = high moisture
content = high K
• Want to know either
– Saturated hydraulic conductivity, Ks, or
– Unsaturated hydraulic conductivity, K.
Other considerations
• What should the sample size be?
• Where to conduct experiment?
• How is the water applied?

• Sample size
– Contemporary soil core devices.
– Representative Elementary Volume (REV).
Experiment location
• Field
• Soil is undisturbed.
• Can’t control the environment.
• Logistics.
• Laboratory
• Highly controlled environment.
• Sample can be aggravated during transport.
• Facilities
Water Application
• Ideally, the soil should be wetted from
the bottom up.
• Should use a deaerated 0.005 M
CaSO4 solution to limit air retention.
• What volume of water is required and
what volume is available.
Determining Ks
• Laboratory Methods
• Field Methods
– Test basins
• Note: for each method….
– good contact must be made at the lateral
boundaries of the core.
– Evaporation must be measured.
• Wet the column from the bottom up.
– Can be a problem depending on sample size.
• Add water until it’s at the desired height.
– Hydraulic gradient = 1 (Figure 10.1a)
– Macropore collapse? Need a different gradient.
– L y x      H    (Figure 10.1b)

L        L

• Capture the outflow, when it’s rate becomes
constant Ks is obtained.
L y x    H

L       L
• L is length through the soil
• y is the height of ponded water
• x is the height of water required to lower
the gradient so that y can be maintained.
• Note: if the gradient is 1 then Ks = q as
per Darcy’s Law.
• Wet the column from the bottom up.
• Fill a burette to above the height of the soil
column and allow it to drain.
• Drain until the rate of head loss is constant.
•           aL              
H2   (Figure 11.1)
Ks                 log H1 
 A(t 2  t 1)   
 aL   H 2 
Ks                 log H1 
 A(t 2  t 1)   

• a is the cross-sectional area of the
burette
• A is the cross-sectional area of the soil
column
• t2 – t1 is the time required for the head
to drop from H1 to H2.
Test Basin Method
• Isolate a column of soil
– Usually much larger than a core to be used
in the laboratory.
•   Seal the lateral faces of the column
•   Ensure the column is saturated
•   Apply a constant head of water at rate P.
•   Obtain Ks using a mass balance
approach: I = P - E where, Ks is equal to I
since the soil is saturated.
Ks Method Summary
• The constant head method is used for
soil with a high Ks (> 0.001 cm/s).
• The falling head method is used for
soils with lower Ks (10-3 - 10-6 cm/s).
• Laboratory experiments can obtain Ks in
each dimension.
Determining Unsaturated K
• Field methods
– Ring infiltrometer.
• Laboratory methods
– Instantaneous profile method.

• Note: ensure that all instruments make
good contact with soil.
Ring Infiltrometer
• Used either in the field or laboratory.
• Can use either one or two rings.
– Scale dependent on ring size.
– 2 rings allows vertical K to be isolated.
• Can measure K when the matric
potential, ym, is > 0.
– When ym is 0 a surface crust of a known
potential can be used.
Ring Infiltrometer Method

• Isolate soil column as in other methods.
• Place the infiltrometer on the soil,
ensuring good contact.
• Water is ponded on the soil and the
infiltration rate recorded.
• Unsaturated K is determined using the
Richards’ equation.
Ring Infiltrometer

Water Supply

Double Ring
Instantaneous Profile Method
• Uses tensiometers and gamma ray
absorption to measure matric potential, f,
and moisture content, q, respectively.
• Pond water until the outflow is constant
and then start the experiment when the
last of the water has entered the soil.
• K is obtained using q   f 
 K 
t z  z 
Instantaneous Profile Method
Tensiometers          TDR’s

Gamma
Ray                          Gamma Ray
Detector                     Emitter
Unsaturated K Method
Summary
• Ring infiltrometer
– Different sample sizes require different
rings and sometimes infiltrometers.
– Water can be hard to provide depending
on the sample size.
– Have to ensure good contact with soil.
• Instantaneous profile method
– Expensive to operate and hard to set up.
– Have to ensure good contact with the soil.
Conclusion/Recommendations

• Methods described allow for determining
K in most settings.
• It’s hard to account for macropore flow.
• There is no method for determining
horizontal K in situ.
• Scales of measure are subject to criticism.
Example
Find the hydraulic conductivity of the sands used in
Darcy’s first series experiments (Refer the Figure),
assuming that the height of the sand column is 3 m and the
diameter of the stand pipe is 0.35 m.

FIGURE: Darcy’s data plotted by Hubbert.
Solution.
Take the flow rate Q = 30 l/min = 0.03 m3/min.
The specific discharge is q = Q/A
= 0.03/(π ∗ 0.352/4)
= 0.312 m/min.
We have the equation,

that yields,       K = qL/h
From the graph     h = 10.5 m.
Thus
K = 0.312 ∗ 3.0/10.5
= 0.089 m/min or 0.0015 m/sec.
This corresponds to a coarse sand.
Example
Find the time it takes for a molecule of water to move from a
factory to a bore hole located 4 km away in a homogeneous
silty sand unconfined aquifer with a hydraulic conductivity of
K = 5 × 10−5 m/sec or 4.32 m/d, an effective porosity of 0.4
and observing that the water table drops 12 m from the factory
to the bore hole.
Solution.
As a simple approximation v = q/ne = Ki/ne
and the pore velocity is calculated as

Since, t = s/v, it would take
time = 4000/(0.0324 ∗ 365) ≈ 338 yr.
If instead, the aquifer was a fractured limestone with a
porosity of 0.01 and the hydraulic conductivity the same,
the pore velocity would be approximately 1.3 m/d and the
time to travel the 4 km would be 8.5 yr.
With a porosity of 0.001 the travel time would reduce to
0.85 yr or about 10 months.
Pumping at the bore hole will increase the hydraulic
gradient and increase the pore velocity and thus decrease
the travel time.
Example
Consider the case of the confined aquifer that is recharged
from an unconfined aquifer through an aquitard. The
recharge rate is 0.3 m/yr or 8.22×10−4 m/d. The water table
is at H = 30 m above the datum. The aquitard is 2 m thick
and its vertical hydraulic conductivity is K’=10−3 m/d. The
unconfined aquifer is 20 m thick and has a hydraulic
conductivity K = 10−1 m/d. Find the piezometric head h at
the bottom of the unconfined aquifer and the difference in
elevation between the water table and the piezometric
surface of the confined aquifer.
Solution.
Let y be the height of the piezometric surface over the top of the aquitard
and z the difference in elevation between the water table and the
piezometric surface (see Figure).
Applying Darcy’s law between points A and B i.e. using the Equation
A        A       B         B
8.22 × 10−4 = 10−1(H − h)/(y + z).
h = H − 8.22 × 10−3(y + z) = 30 − 8.22 × 10−3(20) = 29.84 m.
Writing Darcy’s equation between the top and the bottom of the aquitard
yields
8.22 × 10−4 = 10−3(h − h)/b,
thus
h = h − 8.22 × 10−1b = 29.84 − 8.22 × 10−1 × 2 = 28.20.
Hence,
z = H − h = 30 − 28.20 = 1.80 m.

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 views: 11 posted: 10/7/2012 language: English pages: 31
Description: Drainage Engineering