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.