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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
  – Advection-Dispersion-Equation
• Soil characteristics that determine K
  – Particle size
  – Porosity
  – Bulk density
            More about K
• 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
  – Advantages
     • Soil is undisturbed.
  – Disadvantages
     • Can’t control the environment.
     • Logistics.
• Laboratory
  – Advantages
     • Highly controlled environment.
  – Disadvantages
     • 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
  – Constant head
  – Falling head
• Field Methods
  – Test basins
• Note: for each method….
  – good contact must be made at the lateral
    boundaries of the core.
  – Evaporation must be measured.
       Constant Head Method
• 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.
Constant Head Apparatus
             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.
        Falling Head Method
• 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)   
Falling Head Apparatus
                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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posted:10/7/2012
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Description: Drainage Engineering