Chapter 3 1s2i4

FLOW OF WATER THROUGH SOIL CHAPTER 3

Learning outcome At the end of this lecture, student should be able to: Define the concept of pressure head and hydraulic gradient Understand the definition of permeability and 1D flow of water through soil using Darcy Law Determine the coefficient of permeability using constant head and falling head test

Content  Definition of pressure head and hydraulic gradient  Permeability and 1-D flow of water through soil – Darcy Law  Determination of the coefficient of permeability – constant head, falling head  Field permeability test – confined aquifer, unconfined aquifer  Seepage and 2D-flow in isotropic and homogeneous soil. Seepage calculation using flow net under concrete Dam/sheet piles and through earth dam

INTRODUCTION 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 E.g. 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

INTRODUCTION Water is free to flow within a soil mass In porous media, water will flow from the zones of higher to lower pore pressure When considering problems of water flow, it is usual to express a pressure as a ‘pressure head’ or ‘head’, measured in meter of water

PRESSURE HEAD AND HYDRAULIC GRADIENT  Bernoulli’s equation states 3 heads components, total head (h) causing a water flow

 Where h = total head u = pressure v = velocity g = acceleration due to gravity γw = unit weight of water Z = vertical distance of a given point above or below datum

PRESSURE HEAD AND HYDRAULIC GRADIENT If Bernoulli’s eq is applied to the flow of water through a porous medium, the velocity head can be neglected because the seepage velocity is very small And the total head at any point can be as follows

PRESSURE HEAD AND HYDRAULIC GRADIENT

Figure 1: Pressure, elevation and total heads for flow of water through soil

PRESSURE HEAD AND HYDRAULIC GRADIENT  The loss of head between two points, A and B, can be given by Δh = hA – hB

 The head loss, Δh can be expressed in a nondimensional form as i = Δh L

Where i = hydraulic gradient L = distance between points A and B (the length of flow over which the loss of head occurred)

PRESSURE HEAD AND HYDRAULIC GRADIENT

Figure 2: Nature of variation of v with hydraulic gradient, i

PRESSURE HEAD AND HYDRAULIC GRADIENT  In general, the variation of the velocity v with the hydraulic gradient i is divided into 3 zones i/ laminar flow zone (Zone I) ii/ transition zone (Zone II) iii/ turbulent flow zone (Zone III)  In most soils, the flow of water through the void spaces can be considered laminar Vi

 In fractured rock, stones, gravels and very coarse sands, turbulent flow conditions may exist

COEFFICIENT OF PERMEABILITY Permeability (hydraulic conductivity) – the capacity of a soil to allow water to through it Hydraulic conductivity is generally expressed in cm/sec or m/sec

COEFFICIENT OF PERMEABILITY  The value k is used as a measure of the resistance to flow offered by soil, and affected by several factor: i/ the porosity of the soil ii/ the particle size distribution iii/ the shape and orientation of soil particles iv/ the degree of saturation/presence of air v/ the type of cation and thickness of adsorbed layers associated with clay minerals (if present) vi/ the viscosity of the soil water, which varies with temperature

COEFFICIENT OF PERMEABILITY  Hydraulic conductivity (k) varies widely for different soils  The hydraulic conductivity of unsaturated soils is lower and increases rapidly with the degree of saturation Soil type

k (cm/sec)

k (ft/min)

Clean gravel

100-1.0

200-2.0

Coarse sand

1.0-0.01

2.0-0.02

Fine sand

0.01-0.001

0.02-0.002

Silty clay

0.001-0.00001 0.002-0.00002

Clay

<0.000001

<0.000002

Darcy’s Law In 1856, Darcy published a simple equation for the discharge velocity of water through saturated soils v = ki

Where v = discharge velocity, which the quantity of water flowing in unit time through a unit gross cross-sectional area of soil at right angles to the direction flow k = hydraulic conductivity

Darcy’s Law This equation is on observation about the flow of water through clean sands Valid for laminar flow conditions and applicable for a wide range of soils

EFFECT OF TEMPERATURE The value of the coefficient of permeability will be affected by changes in temperature It may be shown theoretically that for a laminar flow condition in a saturated soil mass: k  γw η

Where γw = unit weight of water η = viscosity of water

EFFECT OF TEMPERATURE A correction for the effect of temperature , may be obtained as follows: kt = kt k20

Where

kt = value k corresponding to a temperature of t k20 = value of k corresponding to a temperature of 20°C kt = temperature correction coefficient

EFFECT OF TEMPERATURE

LABORATORY MEASUREMENTS Coefficient of permeability (k) can be measured using field tests or tests conducted in the laboratory The aim to produce similar results as using field tests In laboratory, errors may occur due to: i/ the presence of air bubbles in the water ii/ Variations in sample density and porosity iii/ variations in temperature and viscosity of water 2 test – constant head test and falling head test

The constant head test

The constant head test To determine the coefficients of permeability (k) of coarse-grained soils such as gravels and sands having value of k above 10-4m/s

The constant head test 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 constant head test Q = Avt = A(ki)t

Where

Q = volume of water collected A = area of cross section of the soil specimen t =duration of water collection q = Av = Aki

Where q = quantity flowing in unit time

The constant head test And because

i=h L

The equation can be substituted as below k = QL Aht

Example 1 During a test using a constant-head permeameter, the following data were collected. Determine the average value k Diameter of sample = 100mm Temperature of water = 17° Distance between manometer tapping points =150mm Quantity collected in 2 min. (ml)

541

503

509

474

Difference in manometer levels (mm)

76

72

68

65

Solution

Solution

Example 2 With the aid of sketch, derive the formula for the permeability of a soil using a constant head apparatus

Solution  By applying Darcy’s equation q = kAi k = q/Ai but q = Q/t, k = 1/A x Q/t x L/H k = QL/Ath (mm/s)

i = h/L

Where, q = flow rate (mm3/s) Q = quantity collected in time (s) = Q (ml) x 10 3 (mm3) A = cross sectional area (mm 2) H = different in manometer levels (mm) L = distance between manometer tapping points (mm)

The falling head test

The falling head test To determine the coefficient of permeability of fine soils For these soils, the rate of flow of water through them is too small to enable accurate measurements using the constant head permeameter

Procedure The test is conducted by filling the standpipe with de-aired water and allowing seepage to take place through the sample The height of water in the standpipe is recorded at several time intervals Test repeated using standpipes of different diameter

The falling head test k = 2.303 aL log10 h1 At h2

 Where

a = cross sectional area of the standpipe A = cross sectional area of the sample t = time interval h1 = initial standpipe reading h2 = final standpipe reading L = length of sample

Procedure Specimen – 100mm diameter undisturbed sample Specimens can also be prepared by compaction in a standard mould A wire mesh and gravel filter is provided at the top and bottom of the sample The base of the cylinder is stood in a water reservoir fitted with a constant-level overflow and the top connected to a glass standpipe of known diameter

Example 3 During a test using falling-head permeameter, the following data were recorded. Determine the average value of k. Diameter of sample = 100mm Length of sample = 150mm

Recorded data Standpipe diameter (mm) 5.00 9.00

12.50

Level in standpipe (mm) Initial, h1

Final, h2

Time interval (s)

1200

800

82

800

400

149

1200

900

177

900

700

169

700

400

368

1200

800

485

800

400

908

Solution Cross sectional area of sample, A = 1002 x π 4 Cross sectional area of standpipe a = d2 x π 4 k = 2.303 aL log10 h1 At h2

Solution Recorded data Standpip e diameter (mm)

Initial, h1

5.00

1200

800

82

0.1761

1.854

800

400

149

0.3010

1.744

1200

900

177

0.1249

1.975

900

700

169

0.1091

1.807

700

400

368

0.2430

1.847

1200

800

485

0.1761

1.959

800

400

908

0.3010

1.789

9.00

12.50

Level in standpipe (mm)

Computed Time interval (s)

log10 h1 h2

k (mm/s) x 10-3

Final, h2

Average k = 1.85 x 10-3 mm/s = 1.85 x 10-6 m/s

Exercise  Question 1 The following data were recorded during a constant-head permeability test: Internal diameter = 75 mm Head lost over a sample length of 180 mm = 247 mm Quantity of water collected in 60 s = 626 ml Calculate the coefficient of permeability for the soil

 Question 2 In a falling-head permeability test the following data were recorded: Internal diameter of permeameter = 75.2 mm Length of sample = 122.0 mm Internal diameter of standpipe = 6.25 mm Initial level in standpipe = 750.0 mm Level in standpipe after 15 min = 247.0 mm Calculate the permeability of the soil

Learning outcome At the end of this lecture, student should be able to: Determine the coefficient of permeability for field permeability test – confined aquifer and unconfined aquifer

FIELD PERMEABILITY TEST Comprehensive multiple-well pumping tests can be expensive to be carry out, but offer a high level of reliability The use of site investigation boreholes can be economically advantageous

FIELD PERMEABILITY TEST Steady state pumping tests Pumping tests involve the measurement of a pumped quantity from a well, together with observations in other wells of the resulting drawdown of the ground level Steady state is achieved when a constant pumping rate, the levels in observation wells are then noted The analysis of the results depends on whether the aquifer is confined or unconfined

Pumping test in a confined aquifer  The average hydraulic conductivity for a confined aquifer can be determined by conducting a pumping test from a well with a perforated casing that penetrates the full depth of the aquifer  The pumping rate must not be high enough to reduce the level in the pumping well below the top of the aquifer  Pumping is continued at a uniform rate q until a steady state is reached  The arrangement of a pumping well and two observation wells is shown here

Pumping test in a confined aquifer

Pumping test in a confined aquifer Water can enter the test well only from the aquifer of thickness H The hydraulic conductivity is given as follows k = q log10 (r1/r2) 2.727H (h1-h2)

Pumping test in a confined aquifer Approximation may be derived from a consideration of the radius of influence (r 0) of the pumping It may be assumed that no drawdown of the piezometric head takes place outside the radius of influence r = ro and h =ho

Example 4 A permeability pumping test was carried out from a well sunk into a confined stratum of dense sand. The arrangement of pumping well and observation wells are shown below. When a steady state was achieved at a pumping rate of 37.4m 3/hr, the following drawdown were observed:

pumping well: d = 4.46m observation well 1: d = 0.42m observation well 2: d = 1.15m a) Calculate a value for the coefficient of permeability of the sand using the observation well data b) Estimate the radius of influence at this pumping rate

Solution a) Observation well data: r1 = 50m r2 = 15m ho = 11.7 + 7.4 – 2.5 = 16.6m h1 = 16.6 – 0.42 = 16.18m h2 = 16.6 – 1.15 = 15.45m q = 37.4 / 3600 = 10.39 x 10-3 m3/s k = q log10 (r1/r2) 2.727 H (h1-h2) = (10.39 x 10-3 ) log 10 (50/15) 2.727 x 11.7 x (16.18 – 15.45) = 2.33 x 10-4 m/s

H = 11.7m

(b) No drawdown Then, putting r1 =50m and h1 = 16.18m k = q log10 (ro/r1) 2.727 H (ho-h1) Log10 (ro/50) = 2.33 x 10-4 x 2.727 x 11.7 16.6 – 16.18 10.39 x 10 -3 ro = 100m

Pumping test in an unconfined aquifer An unconfined aquifer is a free-draining surface layer underlain by an impervious base During the test, water is pumped out at a constant rate from a test well that has a perforated casing Several observations wells at various radial distance are made around the test well Continuous observation of water level in the test well are made after the start of pumping until a steady state is reached

Pumping test in an unconfined aquifer

Pumping test in an unconfined aquifer The hydraulic conductivity is given as follows k = 2.303q log10 (r1/r2) π(h12-h22)

Example 5 A permeability test was carried out from well sunk through a surface layer of medium dense sand. Initially, the water table was located at a depth of 2.5m. When a steady state was achieved at a pumping rate of 23.4m 3/hr, the following draw-downs were observed Pumping well: Observation well 1: Observation well 2:

d = 3.64m d = 0.48m d = 0.96m

(a) Calculate value for the coefficient of permeability of the sand using the observation well data (b) Estimate the radius of influence at this pumping rate

Solution (a) Observation data : r1 = 62m r2 = 18m ho = 12 – 2.5 = 9.5m h1 = 9.5 – 0.48 = 9.02m h2 = 9.5 – 0.96 = 8.54m q = 23.4/3600 = 6.5 x 10-3 m3/s k = 2.303q log10 (r1/r2) π(h12-h22) = 3.04 x 10-4 m/s

(b) Putting r1 =62m and h1 = 9.02m k = 2.303q log10 (ro/r1) π(ho2-h12) ro = 229 m

Exercise  Question 1  For a field pumping test a well was sunk through a horizontal layer of a sand which proved to be 14.4 m thick and to be underlain by a stratum of clay. Two observation wells were sunk, respectively 18 m and 64 m from the pumping well. The water table was initially 2.2 m below the ground level. At a steady state pumping-rate of 328 litres/min, the drawdowns in the observation wells were found to be 1.92 and 1.16 m respectively. Calculate the coefficient of permeability of the sand.   Question 2  A horizontal layer of sand of 6.0 m thickness is overlain by a layer of clay with a horizontal surface thickness of 4.8 m. An impermeable layer underlies the sand. In order to carry out a pumping test, a well was sunk to the bottom of the sand and two observation wells were sunk through clay just into the sand at distances 12 m and 40 m from the pumping well. At a steady pump rate of 600 litres/min., the water levels in the observation wells were reduced by 2.28 m and 1.79 m respectively. Calculate the coefficient of permeability of the sand if the initial piezometric surface level lies 1.0 m below the ground surface.

Seepage and 2-D Flow in Isotropic and Homogeneous Soil  In preceding lesson, we considered some simple cases for which direct application of Darcy’s law was required to calculate the flow of water through soil  In many instances, the flow of water through soil is not in one direction only, nor is it uniform over the entire area perpendicular to the flow  The seepage taking place around sheet-piling, dams, under other water-retaining structures and through embankments and earth dams is two dimensional  Vertical and horizontal velocity components vary from point to point within the cross-section of the soil mass  Graphical representation known as a flow net will be introduced

The flow of water through soils is described by Laplace’s equation.

Where H =total head kx and kz = hydraulic conductivities in X and Y directions Laplace’s eq expresses the condition that the changes of hydraulic gradient in one direction are balanced by changes in the other directions

Laplace’s equation is also called the potential flow equation because the velocity head is neglected If the soil is isotropic with respect to the hydraulic conductivity – that is kx = kz, the preceding continuity equation for 2-D flow simplifies to There are 2 techniques for Laplace’s equation. One of it is an approximate method called flownet sketching

Flow Net A flownet is a graphical representation of a flow field that satisfies Laplace’s equation and comprises a family of flow lines and equipotential lines  Flow line - a line along which a water particle will travel from upstream to the downstream side in the permeable soil medium  Equipotential line - a line along which the potential head at all points is equal

A combination of a number of flow lines and equipotential lines is called a flow net

Completed flow net

Construction of Flow Net  Draw the structure and soil mass to suitable scale  Identify impermeable and permeable boundaries  Sketch a series of flow lines (4 or 5) and then sketch an appropriate number of equipotential lines such that the area between a pair of flow lines and a pair of equipotential lines is approximately a curvilinear square  Theoritically, any no. of flow lines may be drawn and the greater the no., the more accurate should be the calculations that follow.  However, from a practical point of view the task is simplified by drawing only few flow lines; it is not often that more than 5 or 6 will be necessary

Seepage Calculation  In any flow net, the strip between any two adjacent flow lines is called a flow channel.  Let h1, h2, h3, h4,…hn be the piezometric levels corresponding to the equipotential lines  The rate of seepage through the flow channel per unit length (perpendicular to the vertical section through the permeable layer) can be calculated as follows  Δq1=Δq2=Δq3=…..=Δqn

From Darcy’s law, the flow rate is equal to kiA. Thus If the number of flow channels in a flow net is equal to Nf, the total rate of flow through all the channels per unit length can be given by

Or Where H =head difference between the upstream and downstream sides Nd = number of potential drop Nf = number of flow channels in flow net n = b1/l1=b2/l2=b3/l3=……=n elements are not square)

(i.e.

the

EXAMPLE

ANSWER

SUMMARY  In this chapter, we’ve discussed Darcy’s Law, definition of hydraulic conductivity, laboratory and field determination of hydraulic conductivity  The accuracy of the values of k determined in the laboratory depends on several factor - Temperature of the fluid - Viscosity of the fluid - Trapped air bubbles present in the soil specimen - Degree of saturation of the soil specimen - Migration of fines during testing - Duplication of field conditions in the laboratory  The actual value of the hydraulic conductivity in the field may also be somewhat different than that obtained in the laboratory because of the nonhomogeneity of the soil  Hence, proper care should be taken in assessing the order of the magnitude of k for all design consideration.