Flow Of Water Through Soils.docx

ONE-DIMENSIONAL FLOW OF WATER THROUGH SOILS Importance  We have discussed particle sizes and index properties, and used these to classify soils.  You know that water changes the soil states in fine-grained soils; the greater the water content in a soil, the weaker it is.  Soils are porous materials, much like sponges.  Water can flow between the interconnected voids.  Particle sizes and the structural arrangement of the particles influence the rate of flow.  The flow of water has caused instability and failure of many geotechnical structures (e.g., roads, bridges, dams, and excavations).  The key physical property that governs the flow of water in soils is hydraulic conductivity (also called permeability).  A sample practical application is as follows. An excavation is required to construct the basement of a building. During construction, the base of the excavation needs to be free of water. The engineer decides to use a retaining wall around the excavation to keep it dry. Water from outside the excavation will flow under the wall. This can lead to instability as well as flooding of the excavation. To determine the length of the wall to keep the excavation dry, the soil’s hydraulic conductivity must be known. Definitions of Key Terms    

Groundwater is water under gravity that fills the soil pores. Head (H) is the mechanical energy per unit weight. Hydraulic conductivity, sometimes called the coefficient of permeability, (k) is a proportionality constant used to determine the flow velocity of water through soils. Porewater pressure (u) is the pressure of water within the soil pores.

Questions to Guide Your Reading 1. What causes the flow of water through soils? 2. What law describes the flow of water through soils? 3. What is hydraulic conductivity and how is it determined? 4. What are the typical values of hydraulic conductivities for coarse-grained and fine-grained soils? Head and Pressure Variation in a Fluid at Rest We will be discussing gravitational flow of water under a steady-state condition. You may ask: “What is a steady-state condition?” Gravitational flow can only occur if there is a gradient. Flow takes place downhill. The steady-state condition occurs if neither the flow nor the porewater pressure changes with time. Porewater pressure is the water pressure within the voids.

Darcy’s law governs the flow of water through soils. But before we delve into Darcy’s law, we will discuss an important principle in fluid mechanics—Bernoulli’s principle—that is essential in understanding flow through soils. If you cap one end of a tube, fill the tube with water, and then rest it on your table (Figure 6.1), the height of water with reference to your table is called the pressure head (h p). Head refers to the mechanical energy per unit weight. If you raise the tube above the table, the mechanical energy or total head increases. You now have two components of total head—the pressure head (hp) and the elevation head (hz).

Fig.1 Illustration of elevation and pressure heads. If water were to flow through the tube with a velocity v under a steady-state condition, then we would have an additional head due to the velocity, given as v2/2g. The total head, H, according to Bernoulli’s principle, is 𝑯 = 𝒉𝒛 + 𝒉𝒑 +

𝒗𝟐 𝟐𝒈

------------------- (1)

The flow is assumed to be steady, inviscid (no change in viscosity), incompressible (no change in volume), and irrotational (fluid particles do not spin). The elevation or potential head is referenced to an arbitrary datum, and the total head will change depending on the choice of the datum position. Therefore, it is essential that you identify your datum position in solutions to flow problems. Pressures are defined relative to atmospheric pressure (atmospheric pressure is 101.3 kPa at a temperature of 150C). This is called gage pressure. The gage pressure at the groundwater level (free surface) is zero. The velocity of flow through soils is generally small (<1 cm/s) and we usually neglect the velocity head. The total head in soils is then 𝒖 𝑯 = 𝒉𝒛 + 𝒉𝒑 = 𝒉𝒛 + 𝜸 ------------------- (2) 𝒘

Where, u = hp𝛾𝑤 = porewater pressure

Consider a cylinder containing a soil mass with water flowing through it at a constant rate, as depicted in Figure 6.2. If we connect two tubes, A and B, called piezometers, at a distance L apart, the water will rise to different heights in each of the tubes. The height of water in tube B near the exit is lower than that in tube A. Why? As the water flows through the soil, energy is dissipated through friction with the soil particles, resulting in a loss of head. The head loss between A and B, assuming decrease in head, is positive and our datum arbitrarily selected at the top of the cylinder is ΔH = (hp)A - (hp)B. In general, the head loss is the total head at A minus the total head at B.

Fig. 2 Head loss due to flow of water through soil. The ordinary differential equation to describe one-dimensional pressure variation of a fluid at rest (acceleration equal to zero) is 𝒅𝒑 = 𝜸𝒘 -------------- (3) 𝒅𝒛 The fluid pressure difference between two vertical points, z1 and z2, below the free surface (Fig. 3) is 𝑝2

𝑧2

∫ 𝑑𝑝 = 𝛾𝑤 ∫ 𝑧 𝑝1

𝑧1

Fig. Hydrostatic or porewater pressure variation below the groundwater level. Performing the integration gives p2 = p1 = 𝜸𝒘 (z2 - z1) ----------------- (4) At the free surface (z1 = 0), the (gage) pressure is zero (p1 = 0) and z2 = zw, so the fluid pressure variation (called the hydrostatic pressure distribution) is p = u = 𝜸𝒘 zw --------- (5) where, zw = depth from the groundwater level. Porewater pressures are measured by porewater pressure transducers (Fig. 4) or by piezometers (Fig. 5).

Fig. 4 Schematic of a porewater pressure transducer.

Fig. 5 Piezometers. In a porewater pressure transducer, water passes through a porous material and pushes against a metal diaphragm to which a strain gauge is attached. The strain gauge is usually wired into a Wheatstone bridge. The porewater pressure transducer is calibrated by applying known pressures and measuring the electrical voltage output from the Wheatstone bridge. Piezometers are porous tubes that allow the passage of water. In a simple piezometer, you can measure the height of water in the tube from a fixed elevation and then calculate the porewater pressure by multiplying the height of water by the unit weight of water. A borehole cased to a certain depth acts like a piezometer. Modern piezometers are equipped with porewater pressure transducers for electronic reading and data acquisition. Example 1 Determination of Hydraulic Heads Determine (a) the variations of the elevation, pressure, and total heads through the soil when the pressure gage in the experimental setup shown in Figure E6.1a has a pressure of 19.6 kPa, and (b) the elevation, pressure, and total heads in the middle of the soil.

Strategy The first thing you need to do is to define your datum. Outflows or exits are good choices; the pressure head there is zero. The pressure from the pressure gage should be converted to a pressure head by dividing by the unit weight of water (9.8 kN/m3). To determine the pressure head at a point, assume that you connect a small tube at that point and then figure out how high the water will rise in the tube. Solution: Step 1: Define the datum position. Choose C as datum (Figure E6.1b).

Step 2: Determine the heads. Top of soil—B Pressure gage = 19.6 kPa; equivalent pressure head = pressure/unit weight of water = 19.6/9.8 = 2 m = 200 cm Elevation head = 50 cm; pressure head of water above B + pressure gage head = 100 + 200 = 300 cm; total head = elevation head + pressure head = 50 + 300 = 350 cm Bottom of soil—C Elevation head = 0 cm; pressure head 5 0 cm; total head 5 0 1 0 5 0 cm Step 3: Determine the heads at the center of the soil. The heads are linearly distributed through the soil. Therefore, the heads are proportional to the soil height. At the center of the soil, elevation head = 25 cm, pressure head = 300/2 = 150 cm, and the total head = 25 + 150 = 175 cm Darcy’s Law Darcy (1856) proposed that average flow velocity through soils is proportional to the gradient of the total head. The flow in any direction, j, is 𝒅𝑯

𝒗𝒋 = 𝒌𝒋 𝒅𝒙 ------------- (6) 𝒋

Where, v is the average flow velocity, k is a coefficient of proportionality called the hydraulic conductivity (sometimes called the coefficient of permeability), and dH is the change in total head over a distance dx. 𝒗𝒙 = 𝒌𝒙

∆𝑯 𝑳

= 𝒌𝒙 𝒊 ----------- (7)

Where, i = ΔH/L is the hydraulic gradient. Darcy’s law is valid for all soils if the flow is laminar. The average velocity, v, calculated from Eq. (7) is for the cross-sectional area normal to the direction of flow. Flow through soils, however, occurs only through the interconnected voids. The velocity through the void spaces is called seepage velocity (vs) and is obtained by dividing the average velocity by the porosity of the soil: 𝒗𝒔 =

𝒌𝒋 𝒏

𝒊 ----------- (8)

The volume rate of flow, qj, or, simply, flow rate is the product of the average velocity and the cross-sectional area: 𝒒𝒋 = 𝒗𝒋 𝑨 = 𝑨𝒌𝒋 𝒊 ------------ (9)

The unit of measurement for qj is m3/s or cm3/s. The conservation of flow (law of continuity) stipulates that the volume rate of inflow (qj)in into a soil element must equal the volume rate of outflow, (qj)out, or, simply, inflow must equal outflow: (qj)in = (qj)out. Factors Affecting Permeability The hydraulic conductivity depends on 1. Soil type: Coarse-grained soils have higher hydraulic conductivities than fi ne-grained soils. The water in the double layer in fi ne-grained soils significantly reduces the seepage pore space. 2 2 2. Particle size: Hydraulic conductivity depends on 𝐷50 (or 𝐷10 ) for coarse-grained soils.

3. Pore fluid properties, particularly viscosity: k1 : k2 < m2 : m1, where m is dynamic viscosity (dynamic viscosity of water is 1.12 3 1023 N.s/m2 at 15.6 0C) and the subscripts 1 and 2 denote two types of pore fluids in a given soil. 4. Void ratio: k1 : k2 ≈ 𝑒12 : 𝑒22 , where subscripts 1 and 2 denote two types of soil fabric for coarsegrained soils. This ratio is useful in comparing the hydraulic conductivities of similar soils with different void ratios. However, two soils with the same void ratio can have different hydraulic conductivities. 5. Pore size: The greater the interconnected pore space, the higher the hydraulic conductivity. Large pores do not indicate high porosity. The flow of water through soils is related to the square of the pore size, and not the total pore volume. 6. Homogeneity, layering, and fissuring: Water tends to seep quickly through loose layers, through fissures, and along the interface of layered soils. Catastrophic failures can occur from such seepage. 7. Entrapped gases: Entrapped gases tend to reduce the hydraulic conductivity. It is often very difficult to get gas-free soils. Even soils that are under groundwater level and are assumed to be saturated may still have some entrapped gases. 8. Validity of Darcy’s law: Darcy’s law is valid only for laminar flow (Reynolds number less than 2100). Fancher et al. (1933) gave the following criterion for the applicability of Darcy’s law for hydraulic conductivity determination: 𝒗𝑫𝒔 𝜸𝒘 ≤𝟏 𝝁𝒈

Where, v is velocity, Ds is the diameter of a sphere of equivalent volume to the average soil particles, µ is dynamic viscosity of water (1.12 x 10-3 N.s/m2 at 15.6 0C), and g is the acceleration due to gravity. Typical ranges of kz for various soil types are shown in Table 1. TABLE 1 Hydraulic Conductivity for Common Soil Types

Homogeneous clays are practically impervious. Two popular uses of “impervious” clays are in dam construction to curtail the flow of water through the dam and as barriers in landfills to prevent migration of effluent to the surrounding area. Clean sands and gravels are pervious and can be used as drainage materials or soil filters. The values shown in Table 6.1 are useful only to prepare estimates and in preliminary design. Empirical Relationships for k For a homogeneous soil, the hydraulic conductivity depends predominantly on the interconnected pore spaces. You should recall that the pore space (void ratio) is dependent on the soil fabric or structural arrangement of the soil grains. Taylor (1948) proposed a relationship linking k with void ratio as 𝒌𝒛 = 𝑫𝟐𝟓𝟎

𝜸 𝒘 𝑪 𝟏 𝒆𝟑 𝝁 𝟏+𝒆

----------- (11)

Where, C1 is a constant related to shape that can be obtained from laboratory experiments. A number of empirical relationships have been proposed linking kz to void ratio and grain size for coarse-grained soils. Hazen (1930) proposed one of the early relationships as 𝒌𝒛 = 𝑪𝑫𝟐𝟏𝟎 (unit: cm/s) --------- (12) Where, C is a constant varying between 0.4 and 1.4 if the unit of measurement of D10 is mm. Typically, C = 1.0. Hazen’s tests were done on sands with D10 ranging from 0.1 mm to 3 mm and Cu < 5. Other relationships were proposed for coarse-grained and fine-grained soils by Samarasinghe et al. (1982), Kenny et al. (1984), and others. One has to be extremely cautious in

using empirical relationships for kz because it is very sensitive to changes in void ratio, interconnected pore space, and the homogeneity of your soil mass. The Essential Points are: 1. The flow of water through soils is governed by Darcy’s law, which states that the average flow velocity is proportional to the hydraulic gradient. 2. The proportionality coefficient in Darcy’s law is called the hydraulic conductivity, k. 3. The value of kz is influenced by the void ratio, pore size, interconnected pore space, particle size distribution, homogeneity of the soil mass, properties of the pore fluid, and the amount of undissolved gas in the pore fluid. 4. Homogeneous clays are practically impervious, while sands and gravels are pervious. Example 2: Calculating Flow Parameters A soil sample 10 cm in diameter is placed in a tube 1 m long. A constant supply of water is allowed to flow into one end of the soil at A, and the outflow at B is collected by a beaker (Figure E6.2). The average amount of water collected is 1 cm3 for every 10 seconds. The tube is inclined as shown in Figure E6.2. Determine the (a) hydraulic gradient, (b) flow rate, (c) average velocity, (d) seepage velocity if e = 0.6, and (e) hydraulic conductivity.

Strategy In flow problems, you must define a datum position. So your first task is to define the datum position and then find the difference in total head between A and B. Use the head difference to calculate the hydraulic gradient and use Equations (6.7) to (6.9) to solve the problem. Solution: Step 1: Define the datum position. Select the top of the table as the datum. Step 2: Find the total heads at A (inflow) and B (outflow).

Step 3: Find the hydraulic gradient.

EXAMPLE 3: Calculating Hydrostatic Pressures The groundwater level in a soil mass is 2 m below the existing surface. Plot the variation of hydrostatic pressure with depth up to a depth of 10 m. Strategy Since the hydrostatic pressure is linearly related to depth, the distribution will be a straight line starting from the groundwater level, not the surface.

Solution: Step 1: Plot hydrostatic pressure distribution.

Flow Parallel to Soil Layers When the flow is parallel to the soil layers (Figure 6.6), the hydraulic gradient is the same at all points. The flow through the soil mass as a whole is equal to the sum of the flow through each of the layers. There is an analogy here with the flow of electricity through resistors in parallel. If we consider a unit width (in the y direction) of flow and use Equation (6.9), we obtain