2.4 Permeability, Hydraulic Conductivity & Seepage Properties

2.4 Permeability, Hydraulic Conductivity & Seepage Properties

1. Fundamentals of Soil Permeability & Darcy’s Law

Permeability (or hydraulic conductivity, $k$) measures the ease with which a fluid flows through interconnected soil voids. In 1856, Henry Darcy formulated the empirical law governing laminar fluid flow through saturated soils:

$$v = k \cdot i$$ $$Q = v \cdot A = k \cdot i \cdot A = k \cdot \left(\frac{\Delta h}{L}\right) \cdot A$$

Where:

• $q = Q / t$: Volumetric flow rate ($\text{m}^3/\text{s}$ or $\text{cm}^3/\text{s}$).
• $v$: Discharge (superficial) velocity ($\text{m/s}$ or $\text{cm/s}$).
• $k$: Hydraulic conductivity ($\text{m/s}$ or $\text{cm/s}$).
• $i = \frac{\Delta h}{L}$: Hydraulic gradient (head loss $\Delta h$ per length $L$).
• $A$: Total cross-sectional area perpendicular to flow.

Seepage Velocity vs. Discharge Velocity

Discharge velocity $v$ assumes flow across the entire cross-sectional area $A$. However, water flows only through void spaces ($A_v = n \cdot A$). The actual average interstitial fluid velocity, termed seepage velocity ($v_s$), is:

$$v_s = \frac{v}{n} = v \cdot \left(\frac{1 + e}{e}\right)$$

Since porosity $n < 1.0$, seepage velocity is always significantly greater than discharge velocity ($v_s > v$).

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2. Laboratory Measurement of Hydraulic Conductivity

Temperature Correction

Hydraulic conductivity varies inversely with fluid viscosity ($\eta$). Standard values are reported at $20^\circ\text{C}$:

$$k_{20} = k_T \cdot \left(\frac{\eta_T}{\eta_{20}}\right)$$

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3. Empirical Estimations & Field Permeability Testing

Hazen's Formula for Clean Sands

For uniform, clean sands ($D_{10} = 0.1 \text{ to } 3.0 \text{ mm}$):

$$k = C \cdot (D_{10})^2$$

Where $k$ is in $\text{cm/s}$, $D_{10}$ is effective grain size in $\text{mm}$, and $C$ is an empirical coefficient ($1.0 \text{ to } 1.5$).

Pumping Tests in Aquifers (Field Scale)

• Confined Aquifer (Thiem Equation): $$k = \frac{q \ln(r_2 / r_1)}{2 \pi b (h_2 - h_1)}$$ Where $b$ is aquifer thickness, and $h_1, h_2$ are drawdowns at observation wells at radii $r_1, r_2$.
• Unconfined Aquifer (Dupuit Equation): $$k = \frac{q \ln(r_2 / r_1)}{\pi (h_2^2 - h_1^2)}$$

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4. Flow Through Stratified Soil Deposits

Natural sedimentation produces layered soil profiles with anisotropic permeability ($k_h \ne k_v$).

Equivalent Horizontal Hydraulic Conductivity ($k_{h,eq}$)

For flow parallel to soil layering across $n$ horizontal layers of thickness $H_i$ and conductivity $k_i$:

$$k_{h,eq} = \frac{k_1 H_1 + k_2 H_2 + \dots + k_n H_n}{H_{total}} = \frac{\sum k_i H_i}{\sum H_i}$$

Equivalent Vertical Hydraulic Conductivity ($k_{v,eq}$)

For flow perpendicular to soil layering (continuity of velocity $v_1 = v_2 = \dots = v_n$):

$$k_{v,eq} = \frac{H_{total}}{\frac{H_1}{k_1} + \frac{H_2}{k_2} + \dots + \frac{H_n}{k_n}} = \frac{\sum H_i}{\sum \frac{H_i}{k_i}}$$

In all stratified deposits, $k_{h,eq} > k_{v,eq}$. The anisotropy ratio $\frac{k_h}{k_v}$ typically ranges from $2$ to $10+$ in intact varved clays and laminated sands.

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5. Two-Dimensional Steady-State Seepage & Flow Nets

Two-dimensional steady fluid flow through an isotropic porous medium is governed by Laplace's differential equation:

$$\frac{\partial^2 h}{\partial x^2} + \frac{\partial^2 h}{\partial z^2} = 0$$

Flow Net Properties & Rules

A flow net is a graphical solution consisting of two orthogonal families of curves:

1. Flow Lines: Path lines followed by water particles flowing through the soil.
2. Equipotential Lines: Lines connecting points of equal total hydraulic head ($h$).

Construction Criteria:

• Flow lines and equipotential lines intersect at right angles ($90^\circ$).
• Fields formed by intersecting lines form "curvilinear squares" (ratio of mean width to length $b/l \approx 1.0$).
• Impenetrable boundary interfaces (sheet pile, concrete dam base, rock layer) are flow lines or equipotential boundaries.

Quantifying Seepage and Pore Water Pressure

1. Total Volumetric Discharge ($q$):

$$q = k \cdot H \cdot \frac{N_f}{N_d}$$

Where:

• $q$: Flow rate per unit length perpendicular to 2D section ($\text{m}^3/\text{s per m}$).
• $k$: Hydraulic conductivity.
• $H$: Total hydraulic head loss across the structure ($h_{upstream} - h_{downstream}$).
• $N_f$: Number of flow channels.
• $N_d$: Number of equipotential drops.

2. Head Loss per Drop ($\Delta h$):

$$\Delta h = \frac{H}{N_d}$$

3. Pore Water Pressure at any Node $j$:

$$u_j = \left[ h_{upstream} - (N_{drop,j} \cdot \Delta h) - z_j \right] \gamma_w$$

Where $z_j$ is the elevation head of node $j$ relative to datum.

4. Downstream Exit Gradient ($i_{exit}$) and Piping Safety:

$$i_{exit} = \frac{\Delta h}{l_{min}}$$

Where $l_{min}$ is the field length of the smallest square adjacent to the downstream exit face.

$$FS_{piping} = \frac{i_{cr}}{i_{exit}} \ge 3.0$$

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6. Comprehensive Worked Example

Problem Profile: A sheet pile wall penetrates $6.0 \text{ m}$ into a permeable sand stratum ($k = 4.0 \times 10^{-5} \text{ m/s}$) underlain by impermeable clay. The upstream water level is $5.0 \text{ m}$ above the riverbed, while downstream water level is $0.5 \text{ m}$ above the riverbed (Total differential head $H = 4.5 \text{ m}$).

A sketch of the drawn flow net yields:

• Number of flow channels: $N_f = 4$
• Number of equipotential drops: $N_d = 9$
• Field void ratio of sand: $e = 0.60$, $G_s = 2.65$.

Calculate:

1. Total seepage rate per meter length of wall per day ($q$).
2. Head loss per equipotential drop ($\Delta h$).
3. Critical hydraulic gradient ($i_{cr}$) of the sand.
4. Downstream exit gradient ($i_{exit}$) if the smallest field length near exit is $l_{min} = 1.2 \text{ m}$, and factor of safety against piping.

Solution Steps:

1. Calculate Total Seepage Rate ($q$):

   $$q = k \cdot H \cdot \frac{N_f}{N_d} = (4.0 \times 10^{-5} \text{ m/s}) \times (4.5 \text{ m}) \times \frac{4}{9}$$ $$q = 1.80 \times 10^{-4} \times 0.4444 = 8.0 \times 10^{-5} \text{ m}^3/\text{s per meter}$$

   Convert to $\text{m}^3/\text{day per meter}$:

   $$q_{day} = 8.0 \times 10^{-5} \text{ m}^3/\text{s} \times 86,400 \text{ s/day} = 6.912 \text{ m}^3/\text{day per meter}$$

2. Calculate Head Loss per Drop ($\Delta h$):

   $$\Delta h = \frac{H}{N_d} = \frac{4.5 \text{ m}}{9} = 0.50 \text{ m}$$

3. Calculate Critical Hydraulic Gradient ($i_{cr}$):

   $$i_{cr} = \frac{G_s - 1}{1 + e} = \frac{2.65 - 1}{1 + 0.60} = \frac{1.65}{1.60} = 1.031$$

4. Calculate Exit Gradient ($i_{exit}$) and Factor of Safety:

   $$i_{exit} = \frac{\Delta h}{l_{min}} = \frac{0.50 \text{ m}}{1.20 \text{ m}} = 0.4167$$ $$FS_{piping} = \frac{i_{cr}}{i_{exit}} = \frac{1.031}{0.4167} = 2.47$$

Interpretation: The factor of safety ($FS = 2.47$) is below the standard minimum requirement of $3.0$, indicating that additional sheet pile embedment or a downstream filter berm is required to mitigate piping risk.