Soil Permeability and Seepage

Constant Head Permeability Test

• Used for relatively more permeable soils.
• Determines the coefficient of permeability in the laboratory.
• Conducted using a constant-head permeameter.
• Metallic mould dimensions:
  • Internal diameter: $100$ mm
  • Effective height: $127.3$ mm
  • Capacity: $1000$ ml, according to IS: 2720 (Part XVII)
• Includes a detachable extension collar of $100$ mm diameter and $60$ mm height for soil compaction.
• Features a drainage base plate with a recess for a porous stone.
• Fitted with a drainage cap having an inlet valve and an air release valve.
• Drainage base and cap are clamped to the mould.
• Soil sample is placed between two porous discs inside the mould.
• Porous discs should be at least ten times more permeable than the soil.
• Porous discs and water tubes must be de-aired before use.
• Sample preparation involves pouring soil into the permeameter and tamping to achieve the desired density.
• A dummy plate (12 mm thick, 108 mm diameter) is used when compacting the sample in the mould.

Saturation of Soil Sample

• Essential step for accurate testing.
• Methods:
  • Allowing water to flow upward from the base to the top.
  • Pouring soil into the permeameter filled with water, depositing the soil underwater.
  • Applying a vacuum pressure of about $700$ mm of mercury through the drainage cap for about $15$ minutes, followed by saturation with de-aired water from the drainage base.
• Upward flow is maintained until all air is expelled.
• The air-release valve is kept open during the saturation process.

Test Procedure

• After saturation, connect the constant-head reservoir to the drainage cap.

• Maintain a constant water level in the constant-head chamber.

• Allow water to flow from the drainage base until a steady-state is established.

• Collect water spilling over from the constant-head chamber in a graduated jar over a convenient period.

• The head causing flow ($h$) is the difference in water levels between the constant-head reservoir and the constant-head chamber.

• Discharge equation (Eq. 8.3):

  $q = kiA$

  Where:

  • $q$ = discharge
  • $k$ = coefficient of permeability
  • $i$ = hydraulic gradient
  • $A$ = cross-sectional area

• Calculating coefficient of permeability:

  k = rac{qL}{Ah}

  Where:

  • $L$ = length of specimen
  • $h$ = head causing flow

• Discharge ($q$) equals the volume of water collected divided by time.

• Fine soil particles may migrate towards the end faces, forming a filter skin.

• For accurate results, measure head loss ($h'$) over a length ($L'$) in the middle to determine the hydraulic gradient (i = rac{h'}{L'}).

• The temperature of the permeating water should be slightly higher than that of the soil sample to prevent air release.

• A permeameter diameter at least $15$ to $20$ times the particle size reduces void formation at the walls.

• Applying gas pressure to the water surface in the reservoir increases flow rate for low permeability soils. The total head causing flow becomes (h + rac{p}{\gamma_w}), where $p$ is pressure.

• The bulk density of the soil in the mould should match field conditions. Undisturbed samples can be used instead of compacted ones.

• The constant head permeability test is suitable for clean sand and gravel with k > 10^{-2} mm/sec.

Variable-Head Permeability Test

• Used for relatively less permeable soils where the water quantity collected is too small for accurate measurement in the constant-head test.
• Employs the same permeameter mould as the constant-head test.
• A vertical, graduated standpipe of known diameter is fitted to the top of the permeameter.
• The sample is placed between two porous discs, and the assembly is kept in a constant head chamber filled with water.
• Porous discs and water tubes should be de-aired before the sample is placed.
• Undisturbed samples can be used; otherwise, the soil is compacted to the required density in the mould.

Procedure

• The valve at the drainage base is closed, and vacuum pressure is slowly applied through the drainage cap to remove air from the soil.
• Vacuum pressure is increased to $700$ mm of mercury and maintained for about $15$ minutes.
• The sample is saturated by allowing de-aired water to flow upward from the drainage base under vacuum.
• When saturated, both top and bottom outlets are closed, and the standpipe is filled with water to the required height.
• The test starts by allowing water in the standpipe to flow through the sample to the constant-head chamber.
• As water flows, the water level in the standpipe falls. The time required for the water level to fall from an initial head ($h<em>1$) to a final head ($h</em>2$) is recorded.
• The head is measured relative to the water level in the constant-head chamber.

Flow Equations

• At any instant, if the head is $h$, and during a small time interval $dt$, the head falls by $dh$, then:

  $a dh = -q dt$

  Where $a$ is the cross-sectional area of the standpipe.

• Using Darcy's Law:

  $a dh = - (A k i) dt$

  a dh = - A k rac{h}{L} dt

  Where:

  • $A$ = cross-sectional area of the specimen
  • $k$ = coefficient of permeability
  • $L$ = length of the specimen

• Rearranging and integrating:

  $\int<em>{h</em>1}^{h<em>2} \frac{dh}{h} = - \int</em>{t<em>1}^{t</em>2} \frac{Ak}{aL} dt$

  $ln(\frac{h<em>1}{h</em>2}) = \frac{Ak}{aL} (t<em>2 - t</em>1)$

• Solving for $k$:

  $k = \frac{aL}{At} ln(\frac{h<em>1}{h</em>2})$

  Where $t = t<em>2 - t</em>1$ is the time interval.

• Converting to log base 10:

  $k = \frac{2.303 a L}{At} log<em>{10}(\frac{h</em>1}{h_2})$

Discharge Velocity vs. Seepage Velocity

Seepage Velocity

• The discharge velocity ($v$) is a fictitious velocity calculated by dividing the total discharge ($q$) by the total cross-sectional area ($A$).

• The actual velocity through the voids is the seepage velocity ($v_s$), which is greater than the discharge velocity.

• From the continuity of flow:

  $q = vA = v<em>s A</em>v$

  Where $A_v$ is the area of flow through voids.

• Seepage velocity:

  $v<em>s = v \times \frac{A}{A</em>v}$

• Multiplying by length ($L$):

  $v<em>s = v \times \frac{AL}{A</em>v L} = v \times \frac{V}{V_v}$

  Where:

  • $V$ = total volume
  • $V_v$ = volume of voids

• Using porosity ($n = \frac{V_v}{V}$):

  $v_s = \frac{v}{n}$

• Relating to permeability:

  $v<em>s = \frac{ki}{n} = k</em>p i$

  Where $k_p = \frac{k}{n}$ is the coefficient of percolation.

• The coefficient of percolation ($k_p$) is always greater than the coefficient of permeability ($k$).

Permeability of Stratified Soil Deposits

• Natural soil deposits often consist of multiple layers, each with its own coefficient of permeability.
• The average permeability depends on the direction of flow relative to the bedding planes (horizontal, inclined, or vertical).

Flow Parallel to Bedding Planes

• The hydraulic gradient ($i$) is the same for all layers.

• The velocity of flow varies in each layer due to different permeability coefficients.

• Total discharge through the soil deposit is the sum of discharges through individual layers:

  $q = q<em>1 + q</em>2 + … + q_n$

  $k<em>x i Z = k</em>1 i Z<em>1 + k</em>2 i Z<em>2 + … + k</em>n i Z_n$

  Where:

  • $k_x$ = average permeability parallel to bedding plane
  • $Z$ = total thickness of the deposit.

• The average permeability is calculated as:

  $k<em>x = \frac{k</em>1 Z<em>1 + k</em>2 Z<em>2 + … + k</em>n Z_n}{Z}$

Flow Perpendicular to Bedding Planes

• The velocity of flow and unit discharge are the same through each layer.

• The hydraulic gradient and head loss vary in each layer.

• Total head loss is the sum of head losses through individual layers:

  $h = h<em>1 + h</em>2 + … + h_n$

  $i Z = i<em>1 Z</em>1 + i<em>2 Z</em>2 + … + i<em>n Z</em>n$

• If $k_z$ is the average permeability perpendicular to the bedding plane:

  $i = \frac{v}{k_z}$

  $i<em>1 = \frac{v}{k</em>1}, i<em>2 = \frac{v}{k</em>2}, …, i<em>n = \frac{v}{k</em>n}$

• Substituting these values:

  $\frac{vZ}{k<em>z} = \frac{vZ</em>1}{k<em>1} + \frac{vZ</em>2}{k<em>2} + … + \frac{vZ</em>n}{k_n}$

• Solving for $k_z$:

  $k<em>z = \frac{Z}{\frac{Z</em>1}{k<em>1} + \frac{Z</em>2}{k<em>2} + … + \frac{Z</em>n}{k_n}}$

Example Calculation: Stratified Soil Deposit

• A stratified soil deposit has four layers of equal thickness.
• The permeability coefficients of the second, third, and fourth layers are $\frac{3}{2}$, $1$, and $2$ times the permeability of the top layer, respectively.
• Let the thickness of the top layer be $Z$ and its permeability be $k$.

Parallel Permeability Calculation

• $k<em>1 = k, k</em>2 = \frac{3}{2}k, k<em>3 = k, k</em>4 = 2k$
• $\text{Total thickness} = 4Z$
• $k_x = \frac{Z k + Z (\frac{3}{2}k) + Z k + Z (2k)}{4Z}$
• $k_x = \frac{k(1 + \frac{3}{2} + 1 + 2)}{4}$
• $k_x = \frac{11}{8} k$

Perpendicular Permeability Calculation

• $k_z = \frac{4Z}{\frac{Z}{k} + \frac{Z}{\frac{3}{2}k} + \frac{Z}{k} + \frac{Z}{2k}}$
• $k_z = \frac{4}{\frac{1}{k} + \frac{2}{3k} + \frac{1}{k} + \frac{1}{2k}}$
• $k_z = \frac{4}{k^{-1}(1 + \frac{2}{3} + 1 + \frac{1}{2})}$
• $k_z = \frac{4}{\frac{4+8+6+3}{6}} k$
• $k_z = \frac{24}{23} k$

Example Calculation: Coefficient of Permeability

• Soil sample:
  • Height ($L$) = $6$ cm
  • Cross-sectional area ($A$) = $50$ cm$</li> <li>Volume of water passed ($Q$) =$450$ml</li> <li>Time ($t$) =$10$minutes$
  • Effective constant head ($h$) = $40$ cm$</li> <li>Oven-dried weight ($M_d$) =$495$g</li> <li>Specific gravity of soil solids ($G$) =$2.65$</li></ul></li> </ul> <h4 id="permeabilitycalculation">Permeability Calculation</h4> <ul> <li>$k = \frac{QL}{t A h} = \frac{450 \times 6}{10 \times 60 \times 50 \times 40} = \frac{2700}{1200000}$</li> <li>$k = 2.25 \times 10^{-3}$cm/sec</li> <li>$k = 1.944 mm/day$</li> </ul> <h4 id="seepagevelocitycalculation">Seepage Velocity Calculation</h4> <ul> <li>$k = 1.5 \times 10^{-2}$cm/sec (using modified data)</li> <li>Dry density ($\rhod$) =$\frac{Md}{V} = \frac{495}{50 \times 6} = 1.65 g/cm^3$</li> <li>Void ratio ($e$) =$\frac{G \rhow}{\rhod} - 1 = \frac{2.65}{1.65} - 1 = 0.606$</li> <li>Porosity ($n$) =$\frac{e}{1+e} = \frac{0.606}{1+0.606} = 0.377$</li> <li>$v_s = \frac{v}{n} = \frac{1.5 \times 10^{-2}}{0.377} = 3.975 \times 10^{-2}$cm/sec</li> </ul> <h3 id="illustrativeexample81">Illustrative Example 8.1</h3> <ul> <li>Constant head permeameter test observations:<ul> <li>Distance between piezometer tappings =$100$mm</li> <li>Difference of water levels in piezometers =$60$mm</li> <li>Diameter of test sample =$100$mm</li> <li>Quantity of water collected =$350$ml</li> <li>Duration of test =$270$seconds</li></ul></li> </ul> <h4 id="calculation">Calculation</h4> <ul> <li>$q = \frac{350}{270} = 1.296$ml/sec</li> <li>$A = \frac{\pi}{4} (10)^2 = \frac{\pi}{4} (100) = 25 \pi$</li> <li>$k = \frac{qL}{Ah}$</li> <li>$k = \frac{1.296 \times 10}{25 \pi \times 6} = \frac{12.96}{150 \pi} = \frac{0.1296}{15 \pi}$</li> <li>$k= \frac{1.296 \times 10}{ (\pi /4) \times (10)^2 \times 6} = 0.0275$cm/sec</li> </ul> <h3 id="illustrativeexample82">Illustrative Example 8.2</h3> <ul> <li>Falling-head permeability test:<ul> <li>Diameter of soil sample =$4$cm</li> <li>Length of soil sample =$18$cm</li> <li>Head fell from$1.0$m to$0.40$m in$20$minutes</li> <li>Cross-sectional area of standpipe =$1$cm$

Calculation

• $k = \frac{aL}{At} \log<em>e(\frac{h</em>1}{h<em>2}) = \frac{1 \times 18}{(\pi/4) \times (4)^2 \times 20 \times 60} \log</em>e(\frac{1.0}{0.40})$
• $k = \frac{18}{\pi \times 4 \times 20 \times 60} (1.099) = \frac{18}{4800 \pi} (1.099) = 1.30 \times 10^{-3} \hspace{0.1cm}$ cm/sec

Effective Stress, Pore Water Pressure, and Total Stress

Effective Stress

• The pressure transmitted through a soil mass by soil particles at their points of contact.
• Denoted by $\sigma'$, also called intergranular pressure.
• Effective in decreasing void ratio and increasing the shear strength of soil.

Pore Water Pressure

• The pressure transmitted by pore water in a soil mass.
• Denoted by $u$, also referred to as neutral pressure.

Total Stress

• The sum of effective stress and pore water pressure at any point.
• Denoted by $\sigma$
• Formula: $\sigma = \sigma' + u$
• Total stress = Effective stress + Pore water pressure

Capillary Water

Capillary Action

• The ability of a liquid to flow in narrow spaces without external assistance or even against external forces like gravity.
• Surface tension is the tendency of fluid surfaces to shrink into the minimum possible area.

Capillary Water Definition

• Water held in a soil mass due to capillary forces.

Derivation of Expression for Maximum Capillary Rise

• Variables:

  • $d$ = Diameter of the narrow tube
  • $T_s$ = Surface tension (force per unit length)
  • $\alpha$ = Contact angle
  • $h_c$ = Maximum capillary rise
  • $\gamma_w$ = Unit weight of water

• For Vertical Equilibrium:

  $(T<em>s \cos{\alpha}) \pi d = (\pi \frac{d^2}{4})h</em>c \gamma_w$

• Capillary Rise Formula:

  $h<em>c = \frac{4(T</em>s \cos{\alpha})}{d \gamma_w}$

Seepage Analysis

Introduction

• Seepage is the flow of water under gravitational forces in a permeable medium.
• Flow occurs from a point of high head to a point of low head.
• The flow is generally laminar.
• A flow line represents the path taken by a water particle.
• Equipotential lines connect points of equal total head.
• Flow lines and equipotential lines form a flow net, giving a pictorial representation of flow paths and head variation.

Properties of Flow Nets

• Flow lines and equipotential lines meet at right angles.
• Fields are approximately squares, allowing a circle to be drawn touching all four sides.
• The quantity of water flowing through each flow channel is the same.
• The potential drop between two successive equipotential lines is the same.
• Smaller field dimensions indicate a greater hydraulic gradient and velocity of flow.
• In homogeneous soil, transitions in the shape of curves are smooth, being either elliptical or parabolic.