6.2 Flow Nets & Two-Dimensional Seepage Analysis

6.2 Flow Nets & Two-Dimensional Seepage Analysis

Governing Seepage Theory & Laplace Equation

Steady-state two-dimensional fluid flow through homogeneous, incompressible, saturated porous media is governed by Darcy's Law and the Continuity Equation. Assuming laminar flow and rigid soil skeleton, the volumetric water inflow equals outflow for an elemental control volume $(\Delta x \cdot \Delta y \cdot 1)$:

$\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} = 0$

Substituting Darcy's velocities $v_x = -k_x \frac{\partial H}{\partial x}$ and $v_y = -k_y \frac{\partial H}{\partial y}$ yields the general 2D seepage equation:

$k_x \frac{\partial^2 H}{\partial x^2} + k_y \frac{\partial^2 H}{\partial y^2} = 0$

For isotropic soils ($k_x = k_y = k$), this simplifies to Laplace's Differential Equation:

$\frac{\partial^2 H}{\partial x^2} + \frac{\partial^2 H}{\partial y^2} = 0$

Where $H = h_p + h_z = \frac{u}{\gamma_w} + z$ is total hydraulic head. Laplace's equation describes two orthogonal families of curves: Flow Lines (tangent to seepage velocity vectors) and Equipotential Lines (lines connecting points of equal total head $H$).

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Rules of Flow Net Construction & Boundary Conditions

A flow net is a graphical solution of Laplace's equation comprising a grid of flow lines and equipotential lines. Correct manual flow net construction requires strict adherence to physical boundary rules:

                    UPSTREAM HEAD H1
               ~~~~~~~~~~~~~~~~~~~~~~~~~
               |                       |
   FLOW LINE 1 |--------------------   | SHEET PILE WALL
               |                    |  |
               |   EQUIPOTENTIAL    |  |  DOWNSTREAM HEAD H2
               |       DROPS        |  |~~~~~~~~~~~~~~~~~~~~~
               |    +---+---+---+   |  |  |
               |    |   |   |   |   |  |  |  FLOW LINE 4
               |----|---|---|---|---|--|--|------------------>
               |    |   |   |   |   |  |  |
               |====+===+===+===+===|==|==|==================
                         IMPERMEABLE BEDROCK (FLOW LINE)

Boundary Types in Hydraulic Structures:

1. Impermeable Boundaries (Sheet pile surfaces, concrete dam bases, impervious bedrock): Water cannot cross; these surfaces form Flow Lines.
2. Permeable Ground Surfaces (Upstream submerged seabed/riverbed, downstream ground exit): Water enters or leaves soil at uniform total head; these surfaces form Equipotential Lines.
3. Phreatic / Free Water Table Lines (Unconfined seepage through earth dams): Forms the uppermost flow line, along which pore pressure is atmospheric ($u = 0 \implies H = z$).

Core Properties of Isotropic Flow Nets:

• Orthogonality: Flow lines and equipotential lines intersect at exact $90^\circ$ angles everywhere.
• Curvilinear Squares: The ratio of average element width $b$ to element length $l$ in flow direction is constant, usually drawn such that $b/l \approx 1.0$.
• Equal Head Drops: Total head loss $\Delta H$ across the structure is divided into $N_d$ equal equipotential drops: $\Delta h = \frac{\Delta H}{N_d}$
• Equal Discharge per Channel: Total flow rate $q$ is divided into $N_f$ flow channels, each carrying equal discharge $\Delta q$.

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Calculation of Total Seepage Discharge

For a single flow channel element of width $b$ and length $l$, Darcy's Law gives:

$\Delta q = k \cdot i \cdot A = k \cdot \left(\frac{\Delta h}{l}\right) \cdot (b \cdot 1) = k \cdot \Delta h \cdot \left(\frac{b}{l}\right)$

Since $\Delta h = \frac{\Delta H}{N_d}$ and for curvilinear squares $b/l = 1.0$:

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

Summing across all $N_f$ flow channels gives the fundamental Total Seepage Rate Formula per unit length of wall or dam:

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

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Determining Pore Water Pressure and Hydraulic Gradients

To evaluate effective stress, uplifting forces, and stability against piping, total head $H$, elevation head $h_z$, and pore pressure $u$ are computed at any node in the flow net.

1. Total Head ($h_t$) at Node $n$: Counting $n$ equipotential drops from the upstream boundary (where $h_{t,up} = H_1$): $h_{t,n} = H_1 - n \cdot \Delta h = H_1 - n \cdot \left(\frac{\Delta H}{N_d}\right)$

2. Elevation Head ($h_z$): Vertical height of the node above an arbitrary horizontal datum plane ($z = 0$).

3. Pressure Head ($h_p$): $h_p = h_{t,n} - h_z$

4. Pore Water Pressure ($u$): $u = \gamma_w \cdot h_p = \gamma_w \left(h_{t,n} - h_z\right)$

5. Local Hydraulic Gradient ($i_j$): In field element $j$ of length $l_j$: $i_j = \frac{\Delta h}{l_j} = \frac{\Delta H / N_d}{l_j}$

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Seepage in Anisotropic Soil Deposits ($k_x \neq k_y$)

Natural sedimentary soils frequently exhibit horizontal permeability $k_x$ greater than vertical permeability $k_y$ ($k_x / k_y$ typically ranges from 2 to 10). Seepage through anisotropic media is solved by transforming the physical section into an equivalent isotropic section.

Transformation Rules:

1. Horizontal Coordinate Scaling: Scale horizontal physical dimensions $x$ by factor $\sqrt{k_y / k_x}$: $x' = x \cdot \sqrt{\frac{k_y}{k_x}}, \quad y' = y$

2. Equivalent Isotropic Hydraulic Conductivity ($k'$): Use $k'$ in the standard seepage formula: $k' = \sqrt{k_x \cdot k_y}$

3. Seepage Rate Formula for Transformed Section: $q = k' \cdot \Delta H \cdot \frac{N_f}{N_d} = \sqrt{k_x \cdot k_y} \cdot \Delta H \cdot \frac{N_f}{N_d}$

Note on Geometrical Intersections: In the transformed $(x', y')$ drawing, flow lines and equipotential lines intersect at $90^\circ$ forming curvilinear squares. When transferred back to the true scale physical drawing $(x, y)$, lines do not intersect at $90^\circ$ unless $k_x = k_y$.

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Worked Numerical Examples

Worked Example 1: Seepage Discharge & Uplift under Dam

Problem Statement: A concrete dam is founded on an isotropic sand layer ($k = 3.5 \times 10^{-5}\text{ m/s}$). Upstream head $H_1 = 10.0\text{ m}$ above riverbed; downstream head $H_2 = 1.0\text{ m}$ (net head $\Delta H = 9.0\text{ m}$). A flow net has $N_f = 4$ flow channels and $N_d = 12$ equipotential drops. Point A lies on the dam base at elevation $z = -2.5\text{ m}$ relative to downstream riverbed ($z=0$) along the 4th equipotential line (counted from upstream). Calculate:

1. Seepage discharge per meter length per day ($q$).
2. Pore water pressure $u$ at Point A (take $\gamma_w = 9.81\text{ kN/m}^3$).

Solution:

1. Calculate daily seepage rate $q$ per unit length: $q = k \cdot \Delta H \cdot \frac{N_f}{N_d} = (3.5 \times 10^{-5}\text{ m/s}) \times (9.0\text{ m}) \times \left(\frac{4}{12}\right) = 1.05 \times 10^{-4}\text{ m}^3/\text{s/m}$

   Convert to $\text{m}^3/\text{day/m}$ ($86,400\text{ s/day}$): $q = 1.05 \times 10^{-4} \times 86,400 = 9.072\text{ m}^3/\text{day/m}$

2. Calculate pore water pressure at Point A:

   • Head loss per drop: $\Delta h = \frac{\Delta H}{N_d} = \frac{9.0\text{ m}}{12} = 0.75\text{ m}$
   • Total head at upstream riverbed: $H_1 = 10.0\text{ m}$
   • Total head at Point A ($n = 4$ drops): $h_{t,A} = 10.0 - 4 \times (0.75) = 7.0\text{ m}$
   • Elevation head at Point A: $h_z = -2.5\text{ m}$
   • Pressure head at Point A: $h_p = h_{t,A} - h_z = 7.0 - (-2.5) = 9.5\text{ m}$
   • Pore pressure at Point A: $u = \gamma_w \cdot h_p = 9.81\text{ kN/m}^3 \times 9.5\text{ m} = 93.20\text{ kPa}$