CE 35A/L - Geotechnical Engineering I (Soil Mechanics) PDF

# CE 35A/L - GEOTECHNICAL ENGINEERING I (SOIL MECHANICS) ## MODULE 5 ### Permeability and Seepage Prepared by: *Engr. Victor Czar A. Austria* Faculty, CE Department College of Engineering, Architecture and Technology Palawan State University ## PALAWAN STATE UNIVERSITY psu.palawan.edu.ph Thi...

# CE 35A/L - GEOTECHNICAL ENGINEERING I (SOIL MECHANICS) ## MODULE 5 ### Permeability and Seepage Prepared by: *Engr. Victor Czar A. Austria* Faculty, CE Department College of Engineering, Architecture and Technology Palawan State University ## PALAWAN STATE UNIVERSITY psu.palawan.edu.ph This module has the following sections and corresponding icons: | Icon | Section | Description | |---|---|---| | *Image of a book* | Title | It shows the specific topic of the module. | | *Image of a telescope* | Overview | The topics in this module are included in this section. | | *Image of a book with a check mark* | Lecture Proper | A brief debate on the lectures is given in this section. It helps you explore new ideas and capabilities. | | *Image of a brain* | Practice Problems | It involves questions or an expression that sets out the concepts and wordings that you learned from real-life circumstances. | | *Image of a check list* | Assessment | It is a job aimed at evaluating your mastery in acquiring learning skills. | | *Image of a magnifying glass* | Supplementary Knowledge | In this segment you will improve your awareness or experience through the lectures as an additional practice. | | *Image of a key* | Answer Key | This contains answers to all activities in the module. | | *Image of a stack of books* | References | This is a list of all sources that this module uses for creation. | ## OVERVIEW ### COURSE DESCRIPTION: This course deals with soil formation and identification, engineering properties of soils, fundamental aspects of soil characterization and response, including soil mineralogy, soil-water movement, effective stress, consolidation, soil strength, and compaction. Use of soils and geosynthetics in geotechnical and geo-environmental applications. Introduction to site investigation techniques. Laboratory testing and evaluation of soil composition and properties. ### COURSE OUTCOMES: At the end of this module, students will be able to: - Design filters using the permeability and seepage parameters ### INTENDED LEARNING OUTCOMES: At the end of this module, the students will be able to: - Explain terms used in permeability and seepage. - Solve problems on permeability and seepage. - Apply permeability and seepage in filter design. ### PROGRAM OUTCOMES: - An ability to apply knowledge of mathematics, physical sciences, engineering sciences to the practice of civil engineering. - An ability to design and conduct experiments, as well as to analyze and interpret data. - An ability to design, build, improve, and install systems or processes which meet desired needs within realistic constraints. - An ability to recognize, formulate, and solve civil engineering problems. - An understanding of the effects and impact of civil engineering projects on nature. and society, and of the civil engineers' social and ethical responsibilities. - Specialized engineering knowledge in each applicable field, and the ability to apply such knowledge to provide solutions to actual problems . - An ability to effectively communicate orally and in writing using the English language. - An ability to engage in life-long learning and an acceptance of the need to keep current of the development in the specific field of specialization. ### TIME FRAME: This module can be covered in two (2) weeks. ### MODULE OVERVIEW: This module is designed to explain permeability and seepage. It also includes the application of permeability and seepage in filter design. ## LECTURE PROPER ### Lecture 5.1 #### 5.1.1 Introduction Basic Hydrogeology Soils are permeable due to the existence of interconnected voids through which water can flow from points of high energy to points of low energy. The study of the flow of water through permeable soil media is important in soil mechanics. It is necessary for estimating the quantity of underground seepage under various hydraulic conditions, for investigating problems involving the pumping of water for underground construction, and for making stability analyses of earth dams and earth-retaining structures that are subject to seepage forces. #### Importance of Hydrogeology in Geotechnical Engineering: * Moving water affects the properties and behavior of soil and can influence both construction operations and the performance of completed construction. * Permeability and capillary action in soil relates to the type and manner of water movement. * Drainage, seepage, and frost heave relate to practical effects of water movement. * Soil and groundwater contamination expand on environmental effects and concern #### 5.1.2 Hydrologic Cycle The cycle of changes and movements that surface water and shallow groundwater repeatedly pass through. *Image of the hydrologic cycle* ### 5.1.3. Hydrogeology - Is the study of the groundwater phase of the hydrologic cycle (both surface and subsurface) related to the effects of geophysical features of soil and rock formations. #### Precipitation - Snow or rain that reaches the earth's surface tends to either flow over the ground surface as dictated by gravity or to infiltrate into the ground #### Infiltration - Features of nature that influence the rate of infiltration * Infiltrating water that continues to migrate due to gravity effects is drawn to the underground zone. * Slope of the land. * Presence or lack of plant life. * Porosity of soil or rock. #### Water Table or Phreatic Surface - The surface or upper boundary of this saturated-with-water zone, groundwater pressure changes from a negative value to a positive value. - Tends to reflect the surface topography of the area, being at higher elevation where the ground surface is high and at a lower elevation where the ground surface is low. - Depth or elevation varies even on short distances. *Image of the water table* #### Factors affecting the presence of groundwater: * Characteristics of the various soil strata in the soil profile. * Depth and type of underlying rock. * Elevation of the soil surface related to the surrounding terrain. * Surface water coming to the area. * Presence of subsurface water in the adjacent soil and rock deposits. #### Vadose zone - soil zone above the phreatic surface, pore water pressure is negative. #### Groundwater - May be either relatively stationary or mobile (that is, underground flow is occurring) - Conditions for stationary groundwater: * Phreatic surface is at similar elevation over a large area or when adjacent to a large body of water. * Where the top boundary of a subsurface soil or rock layer restricts the entry of water and creates a buried basin in which the collected groundwater is prevented from escaping via movement through the buried soil or rock (perched water table). - Groundwater tends to be mobile where strata of soil or rock have porous characteristics, moving primarily under the effect of gravity. *Image of groundwater cycle* ### Lecture 5.2 #### 5.2.1 Introduction One of the major physical parameters of a soil that controls the rate of seepage through it is hydraulic conductivity, otherwise known as the coefficient of permeability. In this chapter, we will study the following: * Definition of hydraulic conductivity and its magnitude in various soils. * Laboratory determination of hydraulic conductivity. * Empirical relationship to estimate hydraulic conductivity. * Equivalent hydraulic conductivity in stratified soil based on the direction of the flow of water. * Hydraulic conductivity determination from field tests. #### 5.2.2. Bernoulli’s Equation From fluid mechanics, we know that, according to Bernoulli's equation, the total head at a point in water under motion can be given by the sum of the pressure, velocity, and elevation heads. Mathematically, the Bernoulli Theorem expresses the energy head possessed by a body of water at any position as $h = \frac{ν^2}{2g} + \frac{P}{γ} + z$ Where: * h = energy head based on total or absolute hydraulic head, possessed by a body of water at an elevated location (units are length $L$). * v = flow velocity ($L/T$). * g = acceleration of gravity ($L/T^2$). * p = fluid pressure at any point in the body of water ($F/L^2$). * γ = unit weight of water ($F/L^3$). * z = a gravitational position potential head indicating the vertical distance between an arbitrary reference datum and the point where pressure p is measured ($L$). If Bernoulli’s equation is applied to the flow of water through a porous soil medium, the term containing the velocity head can be neglected because the seepage velocity is small, and the total head at any point can be adequately represented by: $h = \frac{P }{γ} + Z $ Where: * $h$ = energy head * $P$ = pressure * $Z$ = elevation #### 5.2.3 Darcy’s Law H. Darcy performed experiments to study the flow of water through sands, it was found that the quantity of water flowing through the soil in a given period was proportional to the soil area normal to the direction of flow and the difference in water levels, and inversely proportional to the length of soil through which flow took place. In 1856, Darcy published a simple equation for the discharge velocity of water through saturated soils, which may be expressed as $v = ki$ Where: * v = discharge velocity, which is the quanity of water flowing in unit time through a unit gross cross-sectional area of soil at right angles to the direction. * k = hydraulic conductivity (otherwise known as the coefficient of permeability) In the Eq. - 5.6, v is the discharge velocity of water based on the gross cross-sectional area of the soil. However, the actual velocity of water (that is, the seepage velocity) through the void spaces is greater than v. #### 5.2.4 Hydraulic Conductivity Hydraulic conductivity is a physical property which measures the ability of the material to transmit fluid through pore spaces and fractures in the presence of an applied hydraulic gradient. The hydraulic conductivity of soils depends on several factors: fluid viscosity, pore size distribution, grain-size distribution, void ratio, roughness of mineral particles, and degree of soil saturation. The value of hydraulic conductivity (k) varies widely for different soils. Some typical values for saturated soils are given in Table 5.1. The hydraulic conductivity of unsaturated soils is lower and increases rapidly with the degree of saturation. | Soil Type | k (cm/sec) | |---|---| | Clean gravel | 100 - 1.0 | | Coarse sand | 1.0-0.01 | | Fine sand | 0.01-0.001 | | Silty clay | 0.001-0.00001 | | Clay | <0.000001 | The hydraulic conductivity of a soil is also related to the properties of the fluid flowing through it by the equation: $k = K = \frac{1}{η_{ad}} = \frac{K}{ρη_{k}}$ Where: * k = coefficient of permeability * γ = unit weight of fluid, kN/m³ * ηad = absolute or dynamic viscosity of the fluid. Pa-sec (1 Pa = 1 N/m²) or poise or centipoise (cp) (1 poise = 1 g/cm·sec or 0.1 Pa·sec); or * ηk = kinematic viscosity, cm²/sec, ft²/sec #### 5.2.5 Permeability, K The ability of water (or other fluid) to flow through a soil by traveling through the void spaces. * Flow is occurring through the void spaces between particles and not actually through the particles themselves. *Image of permeability* The equation for permeability can be expressed as, $K = CSRh^n$ Where: * Cs = shape factor, dimensionless * n = porosity, dimensionless * K = $L^2$ or area * Applicable only to the material through which flow can occur. * It reflects the effect of the size, shape, and number of flow channels, and is independent of any fluid properties. * K, for soil deposits is expressed in darcys (1 darcy = 9.88 x 10^-13 m² or 1.062 x 10^-11 ft²). #### Factors affecting flow: * The pressure difference existing between two points where flow is occurring. * The density and viscosity of the fluid. * The size, shape, and number of pore openings. * The mineralogical, electrochemical or other pertinent properties of the fluid and the soil particles, which affect the attraction between the two materials. #### 5.2.6 Hydraulic Radius, Rh The hydraulic radius is one of the main properties that control the amount of fluid discharge of a channel and its ability to move sediments. This is a measure of channel flow efficiency, is defined as the ratio of the cross-sectional area of fluid flow, A, to the length of the wetted perimeter. $R_h = \frac{Area of Flow}{Wetted Perimeter}$ *Image of hydraulic radius* #### Example 5.1 The relationship expressed by Eq 5.16 permits a soil’s coefficient of permeability (hydraulic conductivity) to be determined with fluids other than water if the value of K, or k for water is known. For example, when compared to water(γ= 9.8 x 10^3 N/m^3, absolute viscosity at 20 °C equal to 1 x 10^-3 Pa·sec), the coefficient of permeability (hydraulic conductivity) for a gasoline (Gg = 0.72 and whose absolute viscosity is 0.337 x 10^-3 Pa-sec) #### Solution: $K_w = (K)_w = (\frac{1}{η_{{ad}_w}}) = k(9.81x20)m/sec$ $k_w = (\frac{ γ_w}{ η_{{ad}_w}}) = k(\frac{9.8x10^3 N/m^3}{1x10^{-3} Pa sec}) = k(9.81x20)m/sec$ $k_g = (\frac{γ_g}{ η_{{ad}_g}}) = k(\frac{72x9.8x10^3 N/m^3}{0.337x10^{-3} Pa sec}) = K(20.9x10)m/sec$ And for the gasoline, $k_g = K(\frac{ γ_g}{ η_{{ad}_g}}) = k(\frac{72x9.8x10^3 N/m^3}{0.337x10^{-3} Pa sec}) = K(20.9x10)m/sec$ And the ratio, $\frac{k_g}{k_w} = \frac{K(20.9x10)}{K(9.8x10)} = 2.1$ ### Lecture 5.3. #### Laboratory Determination of Hydraulic Conductivity Two standard laboratory tests are used to determine the hydraulic conductivity of soil -- the constant-head test and the falling-head test. A brief description of each follow. #### 5.3.1. Constant-Head Test A typical arrangement of the constant-head permeability test is shown in Figure 5.9. In this type of laboratory setup, the water supply at the inlet is adjusted in such a way that the difference of head between the inlet and the outlet remains constant during the test period. After a constant flow rate is established, water is collected in a graduated flask for a known duration. The total volume of water collected may be expressed as $Q = Avt = A(ki)t$ Where: * Q = volume of water collected. * v = discharge velocity. * A = area of cross section of the soil specimen. * t = duration of water collection. And because, $i = \frac{h}{L}$ Where: * L = length of the specimen. Can be substituted to, $Q = A (\frac{kh}{L})t $ Or, $k = \frac{QL}{Aht}$ *Image of the constant-head permeability test* #### 5.3.2 Falling-Head Test A typical arrangement of the falling-head permeability test is shown in Figure 5.10. Water from a standpipe flows through the soil. The initial head difference h₁ at time t=0 is recorded, and water is allowed to flow through the soil specimen such that the final head difference at time t=t2 is h2. The rate of flow of the water through the specimen at any time t can be given by, $q = kA\frac{h}{L}= -a\frac{dh}{dt} $ - Where: - q = flow rate - a = cross-sectional area of the standpipe - A = cross-sectional area of the soil specimen *Image of the falling-head permeability test* Rearrangement of Eq - 5.22 gives, $\frac{dt}{Ak}= a(\frac{-dh}{h})$ Integration of the left side of Eq – 5.23 with limits of time from 0 to t and the right side with limits of head difference from h₁ to h2 gives $\frac{aL}{Ak} t= log_e \frac{h_1}{h_2}$ Or, $k = 2.303 \frac{aL}{At} log_{10} \frac{h_1}{h_2}$ #### Example 5.2 In a laboratory, a constant-head permeability test is performed on a sample of granular soil. The test setup is as indicated in Fig. 5.9. The length of the soil is 15 cm, and the cross-sectional area is 10 cm². If a 24 ml (or 24 cm³) volume of water passes through the soil sample in a 3-minute period, when ∆h is 30 cm, compute the coefficient of permeability. #### Solution: * Q = 24 cm^3 * t = 3 minutes * L = 15 cm * A = 10 cm^2 * Ah = 30 cm $k = \frac{QL}{Aht} = \frac{23(15)}{20(30)(15)}$ $k = 0.4cm/min = 0.067mm/sec$ #### Example 5.3 Refer to the constant-head permeability test arrangement shown in Figure 5.9. Calculate the hydraulic conductivity in cm/sec if a test gives these values: * L = 30 cm * A = 177 cm^2 * h = 50 cm * Water collected in a period of 5 minutes = 350 cm^3 #### Solution: $k= \frac{QL}{Aht} = \frac{350(30)}{177(50)(5)(60 sec/1 min)}$ $k = 0.00395 cm/sec$ #### Example 5.4 For a falling-head permeability test, the following values are given: * Length of specimen = 200 mm * Area of soil specimen = 1000 mm² * Area of standpipe = 40 mm² * Head difference at time t = 0 = 500 mm * Head difference at time t = 180 sec = 300 mm Determine the hydraulic conductivity of the soil in cm/sec. #### Solution: $k = 2.303\frac{aL}{At} log_{10}\frac{h_1}{h_2} = 2.303\frac{40(200)}{1000(180)} log_{10}\frac{500}{300} $ $k= 2.27 × 10^{-2} cm/sec$ #### Example 5.5 A permeable soil layer is underlain by an impervious layer, as shown in the figure. With k = 5.3 × 10^-5 m/sec for the permeable layer, calculate the rate of seepage through it in m³/hr/m width if H = 3m and α = 8°. *Image of a permeable soil layer* #### Solution: * i = $\frac{head loss}{length}$ = $\frac{S tan α}{(S/cosα)}$ = $sin α$ * q = kiA = k(sin α)(3 cos α)(1m) * q = (5.3 × 10−5) (sin 8°) (3 cos 8°) (3600 sec/hr) * q = 0.0789 m³/hr ### Lecture 5.4. ### Directional Variation of Permeability Most soils are not isotropic with respect to permeability. In a given soil deposit, the magnitude of k changes with respect to the direction of flow. #### 5.4. 1 Laminar Flow - Indicates that adjacent paths of water particles are parallel, even when changing direction, and the paths never cross. *Image of a laminar flow* #### 5.4.2 Turbulent Flow - Indicates a disorderly random path for moving water particles with lines of movement crossing and frequently moving at an angle or contrary to the general direction of flow. #### 5.4.2.1 Aquifers in Horizontal Layers (for i ≤ 5, flow is laminar) * i = $\frac{ D1-D2 }{L}$ * α = $\frac{D1 + D2 }{(1)}$ * Flow per unit width, q * q = keqi α *Image of aquifers in horizontal layers* #### 5.4.2.2 Non-homogeneous Unconfined Aquifer * i = $\frac{D1-D2 }{L}$, α = $\frac{D1 + D2 }{(1)}$ * Flow per unit width, q * q = keqi α * Permeability coefficient, k * keq = k1L1+k2L2L *Image of a non-homogeneous unconfined aquifer* #### Example 5.6 Confined aquifer underlies an unconfined aquifer shown in the figure. Given the following: D1 = 59m, D2 = 41m, H1 = 45m, H2 = 33m, K1 = 35m/day, K2 = 27m/day and L = 2m. a. Calculate the coefficient of permeability in horizontal direction. b. Calculate the hydraulic gradient. c. Calculate the flow of water from one stream to another per meter width. *Image of confined and unconfined aquifers* #### Solution: Part a: $k_{eq}H = ΣkH$ $k_{eq}(33+45) = 35(45) +27(33)$ $k_{eq} = 31.615m/day$ Part b: $q = K_{eq}i A$ $i = \frac{h}{L} = \frac{59+41}{2000}$ = $ 0.009$ Part c: $A = \frac{1}{2}(59)(1) + \frac{1}{2}(41) (1) = 50 m^2$ $q = 31.615(0.009)(50) = 14.227 m^3/day$ #### Example 5.7 A layered soil is shown in the figure has the properties of: * H1 = 1.5 m * k1 = 10^-4 cm/sec * H2 = 3 m * k2 = 3.2 x 10^-2 cm/sec * H3 = 2 m * k3 = 4.1 x 10^-5 cm/sec * Estimate the ratio of equivalent hydraulic conductivity, kH/kv . *Image of a layered soil* #### Solution: $(kH)_{eq}= Σ(\frac{k_1H1 +k_2H2 +....k_nH_n}{H}) $ $(kH)_ {eq} = (1.5+3+2)[(10^{-4})(1.5)+(3.2×10^{-2})(3)+(4.1×10^{-5})(2)]$ $(kH)_ {eq} = 148.05 × 10^{-4 } cm/s$ $(k_v) _ {eq}= \frac{H1}{k1}+\frac{H2}{k2}+...+\frac{Hn}{k_n}$ $(k_v) _ {eq}= \frac{1.5}{10^{-4}} +\frac{3}{3.2×10^{-2}} +\frac{2}{4.1×10^{-5}}$ $(k_v) _ {eq} = 1.018 × 10^{-4} cm/s$ The ratio of equivalent hydraulic conductivity = $\frac{(kH)_ {eq}}{(k_v) _ {eq}}= \frac{148.05 × 10^{-4} cm/s}{1.018 × 10^{-4} cm/s} = 145.4$ #### Example 5.8 A confined aquifer has a source of recharge as shown in the figure. The hydraulic conductivity of the aquifer is 40 m/day and its porosity is 0.25. The piezometric head in two wells 1000m apart is 65m and 60m respectively. The average thickness of the aquifer is 25m and its average width is 4km. a. Compute the rate of flow through the aquifer in m³/day. b. Compute the seepage velocity. c. Compute the time of travel from the head of aquifer to a point 3 km from the downstream in days. *Image of a confined aquifer* #### Solution: Part a: * i = $\frac{H}{L} = \frac{65-60}{1000}$ = 5 x 10^-3 * Q = kiA = (5 × 10−3)(40)(25)(4000) * Q = 20,000 m³/day Part b: * V = $\frac{Q}{A}$ = $\frac{Q}{n}$ therefore V = $\frac{Q}{nA}$ * $V_s = \frac{20,000}{(25 x 4,000) 0.25}$ = 0.80m/day Part c: * t = $\frac{Distance} {Seepage Velocity} = \frac{3 x 1,000}{0.80}$ * t = 3750 days #### Example 5.9 Figure below shows three layers of soil in a tube that is 100 mm × 100 mm in cross section. Water is supplied to maintain a constant-head difference of 300 mm cross the sample. The hydraulic conductivities of the soils in the direction of flow through them are as follows. *Image of the soil in a tube* #### Solution: $(kv)_{eq} = Σ(\frac{h_1}{k1}+\frac{h_2}{k2}+...+\frac{h_n}{k_n}) = \frac{150}{10^{-2}} + \frac{150}{3 × 10^{-3}} + \frac{150}{4.9 × 10^{-4}}= 0.001213 cm/sec $ $q = (kv)_{eq}iA = (0.001213) (300) (100mm×100mm) $ $q = 0.0809 cm³/sec = 291.24 cm³/hr$ ### Lecture 5.5. ### Effects of Soil Type and Empirical Relationship - Coarser soils have greater values of k than fine-grained soils. - The variation in flow is caused by the friction developed at the conduit wall and the viscous friction developed in the moving fluid. - For fine-grained soils: lines of flow are physically close to the "wall of the conduit," and therefore only low velocity flows occur - For clays: flow is further hampered because some of the water in the voids is held, or adsorbed, to the clay particles, reducing the flow area and further restricting flow. | Soil Type | Relative Degree of Permeability | k, Coefficient of Permeability or Hydraulic Conductivity (mm/sec) | Drainage Properties | |---|---|---|---| | Clean gravel | High | 10 to 100 | Good | | Clean sand, sand, and gravel Mixtures | Medium | 10 to 10^-2 | Good | | Fine sands, silts | Low | 10^-2 to 10^-4 | Fair-poor | | Sand-silt-clay Mixtures, Glacial tills | Very low | 10^-3 to 10^-6 | Poor-practically impervious | | Homogeneous Clays | Very low-practically impermeable| < 10^-6 | Practically impervious | **Note: To convert, use 1 mm/sec = 0.2 ft/min = 86.4 m/day.** #### 5.5.2 Empirical Relationships ##### 5.5.2.1. For uniform sands in relatively loose condition $k = (D10^2)(10^5)$ * k - permeability rate in mm/sec * D10- 10 percent particle size in mm ##### 5.5.2.2. For dense or compacted sands $k = 3.5 (D15)^2$ * k - permeability rate in mm/sec * D15 - 15 percent particle size in mm ##### 5.5.2.3. For sands where Cu ≤ 10 $k = (1.5 to 3)(3.5) (D15)^2$ * k - permeability rate in mm/sec * D15 - 15 percent particle size in mm ##### 5.5.2.4. For loose sands where Cu > 10 $k = (5 to 15) (3.5) (D15)^2$ * k - permeability rate in mm/sec * D15 - 15 percent particle size in mm ### Lecture 5.6. ### Drainage - The action or method of removing subsurface water by means of channels, pipes or geosynthetics to render the area or structure useful. #### Purposes: * To handle subsurface water encountered during construction sequence. * To handle subsurface water after construction so that the completed facility is not damaged nor its usefulness impaired. #### Benefits: * **During construction:** - Better working conditions for workers and equipment. - Proper construction can be achieved. * **After construction:** - Protection of structure from erosive effects of flowing groundwater like rendering foundation unstable. - Maintaining usability of interior portions of structure located below groundwater elevation. #### 5.6.2 Drainage During Construction - It is understood that excavations extend below water table thus dewatering procedures will be required to keep the work area dry. #### Dewatering - The procedure used to remove water from construction area, such as pumping from an excavation or location where water covers the planned working surface; the procedure used to lower the groundwater table to obtain a "dry” area in the vicinity of an excavation that would otherwise extend below water. ##### a. Dewatering Shallow Excavations * Type of soil: coarse. * Method: open drainage or interceptor ditches. *Image of open drainage of interceptor ditches* #### Drawdown - The lowering of the level of the groundwater table that occurs in the vicinity of a water well (on dewatering equipment) when it is pumped. #### Sump - Small excavation or pit provided in the floor of the structure, or in the earth, to serve as a collection basin for the surface water and near-surface underground water. - Soil strength is reduced as subsurface water flows in an upward direction into an excavation area being dewatered because of seepage force. - Change in strength should be considered in designing excavation bracing and foundations. ##### b. Dewatering Intermediate Depths * Type of soil: coarse grained to silty. * Depth: more than 1 meter or a few feet below groundwater level. ###### Methods employed: * **b.1. Well points:** The perforated end section of a well pipe that permits the groundwater to be drawn into the pipe for pumping. - Depth: 9 meters - Installation: in drilled holes or by jetting. *Image of a well point system or Dewatering* ##### Procedure by jetting: * Water is pumped through the riser and discharges through special openings at the tip of the well point. * Soil below is displaced by jetting action at the well point. * This is continued until the desired depth is achieved. For dewatering purposes: * A series of well points is installed around the perimeter of the area. * Spacing of well points is according to the soil type and depth of dewatering (1 - 3 meters). * Each well riser (50 mm $) is connected by hose or pipe to a horizontal header or collector pipe (150 – 300 mm $). ###### Screen - Serves as protection or filter to prevent soil particles from clogging the well point perforations. - For medium sand and coarser materials surrounding the well point; screen is provided. - For finer soils; sand filter is necessary. ###### Benefits of a sand filter: * It acts to increase the size of the well point, permitting easier and greater flow into the well point. ###### Factors affecting the horizontal extent to which the surrounding soil will show the lowering of the groundwater table : * Depth of the well point * Permeability of the soil as permeability increases so is the horizontal distance ###### Factors affecting the depth of dewatering: * Type of pumping equipment. * Proper assembly of the piping system. * In a single line of well points, the depth of dewatering is usually about 6 – 7 meters. * For lowering the water table to a greater depth, two (or more) – stage (multistage) installation may be required ##### b.2. Vacuum dewatering - The application of vacuum to the piping system to satisfactorily dewater silty soils because capillary forces tend to hold the pore water. * Type of soil: silt. * Height of lift: 5 – 6 meters. *Image of a vacuum dewatering system* ###### To attain vacuum: - The vacuum dewatering system requires that the well point and the riser be surrounded to within a few feet of the ground surface with filter sand. - Then the top few feet be sealed or capped with an impervious soil or other material. - With the pump maintaining a vacuum pressure, the hydraulic gradient for flow to the well points is increased. **Multistage dewatering is necessary due to limited height of lift for deeper excavation.** ##### b.3. Electro – Osmosis * Type of soil: fine silts, clay, or coarse-grained-fine-grained mixtures >25% (+ 0.002 mm) ###### Basic principle: A well or well point is made the cathode, pore water will then migrate toward the negative electrode (cathode). Collected water drained from the soil can be removed by pumping as with the conventional well-point system. ###### Explanation: Groundwater has cations (positive ions) which are the dissolved minerals. The cations concentrate around the negatively charged surface of clay particles to satisfy the electrical charge in the particle. These cations, in turn, attract the negative "end" of dipole water molecules. As the cations are drawn to the cathode, water molecules held to the cation follow. ###### Installation: Anode should be located close to the excavation perimeter and place the cathode farther back. ###### Explanation: The direction of drainage moves away from the excavation thus eliminating the danger that seepage forces may cause like sliding or sloughing in of the exposed slopes toward the excavation. ###### Disadvantages: * It requires special equipment. * High electricity consumption. * Expensive. #### c. Deep Drainage - Are conditions where excavations extend deep below groundwater table or penetrate through a deep permeable stratum. - Deep wells or deep well pumps are used * Deep wells are frequently large diameter (600 mm) drilled