Seepage through Soils

A flow net is a graphical representation of two-dimensional steady seepage, made of two families of curves that cross at right angles: flow lines and equipotential lines. It gives the discharge, pore pressures and exit gradients directly.

• Flow line (stream line): the path followed by a water particle through the soil from upstream to downstream. No water crosses a flow line.
• Equipotential line: a line joining points having the same total head. Water level in a piezometer is the same at all points of the line.

     ==========  upstream
     |  |  |  |    flow lines  ----->
  ---+--+--+--+---  equipotentials  |

Properties

1. Flow lines and equipotential lines intersect at right angles.
2. The fields (elements) are approximately squares, so that a circle can be drawn touching the four sides of each.
3. The same quantity of flow takes place in each flow channel ($\Delta q$ is equal).
4. The head drop between two adjacent equipotential lines is the same ($\Delta h = h/N_d$).
5. Flow lines never cross each other; equipotential lines never cross each other.
6. The smaller the field, the greater the velocity and gradient.
7. Boundaries: an impervious surface is a flow line; the upstream and downstream ground surfaces under water are equipotentials.

Applications

• Seepage discharge: $q = k\,h\,\dfrac{N_f}{N_d}$
• Pore water pressure at any point (for stability and uplift on dams, weirs)
• Exit gradient and check against piping
• Seepage force and uplift pressure under hydraulic structures
• Design of the length of sheet piles, filters and drains
• Location of the phreatic line in an earth dam.

Statement

At every point of a flow net for steady 2D seepage, the flow line and the equipotential line passing through the point are perpendicular.

Proof

Let the head be $h(x,z)$. By Darcy's law (isotropic soil), the velocity components are

Define the velocity potential $\phi = -k\,h$ so $v_x = \dfrac{\partial\phi}{\partial x}$, $v_z = \dfrac{\partial\phi}{\partial z}$. Equipotential lines are $\phi =$ constant.

Slope of the equipotential line: along it $d\phi = 0$:

Slope of the flow line: the velocity vector is tangent to the flow line, so

Product of the slopes:

Since the product of the slopes is $-1$, the two lines are at right angles. $\blacksquare$

Physical reasoning: along an equipotential line there is no head difference, hence no driving gradient and no flow component along it. The velocity (flow direction) is therefore wholly normal to it.

A protective filter is a layer of graded granular material placed between the base soil (the soil being protected) and a drain or coarse material, to allow water to escape without letting soil particles wash out.

Basic requirements

1. Retention (piping) criterion: the filter pores must be small enough to hold back the particles of the base soil.
2. Permeability criterion: the filter must be much more permeable than the base soil, so that no pore pressure builds up and the water escapes freely.
3. The filter must be well graded, with no fines that could migrate, and be stable (no segregation) and durable.
4. It must be thick enough, and not clog; in multiple layers, each layer must satisfy the criteria with respect to the one next to it.

Terzaghi's filter rules for controlling piping

Additional (U.S. Bureau / Corps of Engineers) conditions: $D_{50}(\text{filter})/D_{50}(\text{soil}) < 25$ and the grain-size curve of the filter is roughly parallel to that of the soil. For slotted pipes, $D_{85}(\text{filter}) >$ slot width (or 1.2 times slot width).

Consider a flow net with $N_f$ flow channels and $N_d$ equipotential drops, and a total head loss $h$ across the earth mass.

   equipotential 0     1     2   ...  Nd
        |      |      |      |
 ------ +------+------+------+------ flow line
        |  a   |  a   |  a   |   (a = b : square field)
 ------ +------+------+------+------
          b

Derivation

Take one field of length $a$ (in the direction of flow) and width $b$ (normal to the flow), per unit thickness perpendicular to the section.

Head drop across each field:

Hydraulic gradient across the field:

Discharge through the channel, by Darcy's law ($A = b \times 1$):

For square fields, $a = b$:

The same $\Delta q$ flows in each of the $N_f$ channels, so the total discharge per unit length is

Assumptions: the soil is homogeneous, isotropic and saturated; water and soil grains are incompressible; flow is steady, laminar and obeys Darcy's law; flow takes place in the $x$–$z$ plane.

Consider an elemental soil block $dx \times dz$ with unit thickness. Let the discharge velocity components be $v_x$ and $v_z$.

          v_z + (∂v_z/∂z)dz
              ↑
      +---------------+
 v_x →|    dx × dz    |→ v_x + (∂v_x/∂x)dx
      +---------------+
              ↑
             v_z

Continuity

Steady flow with incompressible water: inflow = outflow.

Darcy's law

(the sign depends on the direction chosen for $h$; it does not affect the result).

Substituting (2) in (1):

For isotropic soil $k_x = k_z$:

This is the Laplace equation. Its solutions represent two families of curves, equipotential lines $h =$ constant and flow lines, which form the flow net.

Piping is the progressive removal of soil particles by seepage water, forming a pipe-like channel beneath a hydraulic structure.

Mechanism:

1. Water seeps under the structure (dam, weir, sheet pile) from the upstream to the downstream side. The flow is concentrated near the downstream toe, where the flow lines converge and the exit gradient is highest.
2. When the exit gradient $i_e$ reaches the critical gradient $i_c = \dfrac{G-1}{1+e} \approx 1$, the upward seepage force cancels the submerged weight of the soil. Effective stress becomes zero and soil grains are lifted and boil out ("boiling").
3. The loss of soil creates a channel at the downstream end that grows backward (towards the upstream) because the shorter flow path increases the gradient. This channel eventually reaches the reservoir, and the structure fails by undermining.

Safety is checked by $F = i_c/i_e \ge 4$ to 5 (or the creep length methods of Bligh and Lane). It is prevented by longer cut-offs, sheet piles, downstream filters and blankets.

A filter (drain) is provided on the downstream side of an earth dam because

• it lets the seepage water leave the dam body freely while holding back soil particles, so piping is prevented;
• it keeps the phreatic line (top flow line) inside the dam and away from the downstream face, so the slope does not become saturated and slump or erode;
• it reduces pore water pressure in the downstream shell, which increases the stability against slope failure;
• it collects the seepage and drains it safely, avoiding softening and sloughing of the downstream toe.

Confined flow: seepage in which all the boundaries of the flow region are fixed and known, so the flow lines and the flow net can be drawn directly. Examples: flow beneath a concrete dam or weir, flow under a sheet pile wall, flow through a pervious foundation.

Unconfined flow: seepage in which one boundary of the flow region, the upper flow line (phreatic line or line of seepage), is not fixed and must be found by trial, because it is a free surface at atmospheric pressure. Example: seepage through the body of an earth dam, or flow to a well in an unconfined aquifer.

The flow through the body of an earth dam is unconfined flow, because the top flow line (phreatic line) is a free surface under atmospheric pressure whose position is not known in advance and must be located.

Flow net under a sheet pile

A flow net for seepage under a sheet pile wall (confined flow) is drawn by trial and error.

 upstream ~~~~~|S|~~~~~ downstream
 h1 ▽__________| |__________▽ h2
   ..flow lines..|..flow lines..
 ----equipotentials cross at 90°-----
 =========== impervious layer ===========

Steps:

1. Draw the section to a convenient scale: sheet pile, ground levels, water levels and the impervious base.
2. Identify the boundary conditions. The upstream soil surface (bed) and the downstream surface are equipotential lines. The surface of the sheet pile and the impervious base are flow lines.
3. Sketch 3 to 4 flow lines, smooth curves starting perpendicular to the upstream equipotential, bending round the pile tip, and ending perpendicular to the downstream equipotential. The line nearest the pile follows the pile; the lowest follows the impervious layer.
4. Sketch equipotential lines to cut the flow lines at right angles so that each field is nearly a square (a circle can be inscribed in each).
5. Adjust the lines by trial until all fields are curvilinear squares.
6. Count $N_f$ (flow channels) and $N_d$ (equipotential drops). Discharge: $q = k\,h\,\dfrac{N_f}{N_d}$ per metre length, with $h = h_1 - h_2$.

Graded filter design

A filter between the base soil and a coarse drain lets water pass but holds back soil particles (piping prevention). For a big grain-size jump, several layers of increasing size are used, each protecting the previous one.

 base soil | filter 1 | filter 2 | coarse drain
  fine     |  finer   | coarser  | stone/pipe
 -------------------------------------------->
          direction of seepage

Terzaghi's criteria, for each filter and the layer it protects:

• Piping: $\dfrac{D_{15}(\text{filter})}{D_{85}(\text{soil})} \le 4 \text{ to } 5$
• Permeability: $\dfrac{D_{15}(\text{filter})}{D_{15}(\text{soil})} \ge 4 \text{ to } 5$
• $\dfrac{D_{50}(\text{filter})}{D_{50}(\text{soil})} \le 25$, and the grading curve of the filter roughly parallel to that of the soil.

Procedure: plot the grading curve of the base soil; find $D_{85}$ and $D_{15}$; fix the grading band of the first layer from the criteria; repeat with the first layer as the "soil" for the next layer until the last layer is coarse enough for the drain. Each layer should have a minimum thickness (about 15 to 30 cm) and contain few fines.

A sheet pile wall is driven into a pervious layer with an impervious base. The water level is higher upstream (head $h_1$) than downstream ($h_2$), so water flows under the pile tip.

  upstream ▽      |P|      ▽ downstream
  ----------------|P|----------------
   flow lines ~~> |P| ~~> curve under tip
  ================================== impervious

Steps to draw the flow net

1. Draw the section to scale showing the sheet pile, soil surfaces and the impervious base.
2. Note boundary conditions: upstream and downstream soil surfaces are equipotential lines; the sheet pile faces and the impervious base are flow lines.
3. Draw 3 or 4 flow lines as smooth curves round the pile tip, perpendicular to the equipotentials.
4. Draw equipotential lines, perpendicular to flow lines, so that the fields are squares.
5. Adjust by trial and error until all fields are curvilinear squares.
6. Count $N_f$ and $N_d$.

Discharge

where $H = h_1 - h_2$ is the net head loss, $N_f$ the number of flow channels and $N_d$ the number of equipotential drops. $q$ is in m³/s per metre length of the wall.

Given

$k = 4\times10^{-2}$ cm/s $= 4\times10^{-4}$ m/s. Permeable layer depth $T = 10$ m, sheet pile penetration $s = 7$ m (so $s/T = 0.7$), net head $H = 3$ m.

Flow net

  H=3m ▽       |P|
  ~~~~~~~~~~~~~|P|~~~~~~ (downstream ground)
   ground      |P|   7 m
   3 flow channels, 8 equipotential drops
   ______________|__|________
   ======== impervious, 10 m deep ========

Draw the flow net with the pile tip 3 m above the impervious layer. Sketching it gives about

• $N_f = 3$ flow channels
• $N_d = 8$ equipotential drops

Seepage quantity

A check with the exact solution for a single sheet pile of penetration ratio $s/T = 0.7$ gives $q/(kH) = 0.371$, i.e. $q = 4.45\times10^{-4}$ m³/s per m, so the flow net values agree.

Answer: $q \approx 4.5\times10^{-4}$ m³/s per metre length ($\approx 39$ m³/day per m).

Given

$q = 200$ cm³/day (per unit width, taken as 1 cm), $N_f = 5$, $N_d = 10$, $h = 2.5$ m $= 250$ cm.

Calculation

Answer: $k = 1.6$ cm/day $\approx 1.85\times10^{-5}$ cm/s.

The top flow line (phreatic line) of an earth dam with a horizontal filter is found by Casagrande's method, using the fact that the line is nearly a parabola (the "base parabola") whose focus is the start of the filter.

          A  B            crest
     ▽ ___ . .____________
      /     ·  \ ·         \
     /  entry   ·· parabola \
    /  correction   ··· ·    \
   /____________________F=====\_ filter
   upstream toe        d     filter

Procedure

1. Draw the dam section to scale. Mark the reservoir water level and point A, where it meets the upstream slope.
2. Mark point B on the water surface at a horizontal distance $0.3\,\Delta$ from A towards the dam, where $\Delta$ is the horizontal projection of the wetted upstream slope. B is the start of the base parabola.
3. The focus F of the parabola is the upstream end of the horizontal filter. Let $d$ be the horizontal distance from B to F and $h$ the height of B above the base.
4. Find the parameter $y_0$ (distance from focus to directrix):

5. The parabola equation (origin at F, $x$ positive upstream) is

Compute points for several values of $y$ and plot the base parabola from B to F. The vertex is at a distance $y_0/2$ from F. 6. Apply the entry correction: draw the top flow line from A, starting perpendicular to the upstream slope (an equipotential), merging smoothly into the parabola by hand. 7. At the filter, the line ends meeting the filter at right angles, with the parabola passing through the focus.

Discharge

The top flow line is the boundary of the unconfined flow; once located, the flow net is completed below it.

Method

Casagrande's base parabola for a dam with a horizontal filter. Seepage per unit length: $q = k\,y_0$ where $y_0 = \sqrt{d^2 + h^2} - d$.

Geometry from the figure (some labels were unclear, so values are deduced)

• Slopes 3H:1V. Water height $h = 30$ m, so the wetted upstream slope has horizontal projection $\Delta = 3\times30 = 90$ m.
• Base width 200 m = $3H_d + 8 + 3H_d$ gives dam height $H_d = 32$ m.
• Point B is $0.3\Delta = 0.3\times90 = 27$ m from A, towards the dam (matches the 27 m in the figure).
• Filter length 45 m at the downstream toe, so the focus F is at $200 - 45 = 155$ m from the upstream toe. B is at $90 + 27 = 117$ m from the upstream toe.

Base parabola parameter

Seepage

$k = 3\times10^{-3}$ mm/s $= 3\times10^{-6}$ m/s

Answer: $q \approx 3.1\times10^{-5}$ m³/s per metre ($\approx 2.7$ m³/day per m).

Meaning of focal length: the Casagrande base parabola has its focus F at the beginning of the horizontal filter. Let $y_0$ be the distance from the focus to the directrix (the "focal length" in this context, the parabola's parameter). The distance from focus to the vertex is $y_0/2$.

   y
   |   .  .           directrix
   |        .   .  <--- parabola
   |   y0 --->   .
   +------F-----------------> x   (filter at the toe)

Proof

Take the origin at F, $x$ positive towards the upstream side and $y$ vertical. Any point P(x, y) on the parabola is equidistant from the focus and from the directrix ($x = -y_0$):

Differentiate with respect to $x$:

By Dupuit's assumption, the hydraulic gradient on a vertical section at height $y$ is $i = dy/dx$ (slope of the free surface), and the flow area per unit width is $y \times 1$:

Hence the discharge per unit width is equal to the coefficient of permeability times the focal length $y_0$. $\blacksquare$

At the focus ($x = 0$) the ordinate is $y = y_0$ and the slope is 1, so $q = k \times 1 \times y_0$, which agrees.

(If the focal length is taken as the focus-to-vertex distance $a = y_0/2$, then $q = 2ka$.)

Given

Top width 15 m; slopes 2H:1V both sides; height 30 m; free board 5 m, so reservoir depth $h = 25$ m; horizontal toe drain 30 m long; $k = 40$ m/day.

Geometry (horizontal drain, Casagrande)

• Base width: $2(2\times30) + 15 = 135$ m.
• Wetted upstream slope has horizontal projection $\Delta = 2\times25 = 50$ m, so A is 50 m from the upstream toe.
• B is $0.3\Delta = 15$ m inside from A, i.e. 65 m from the upstream toe.
• The focus F (start of the drain) is at $135 - 30 = 105$ m from the upstream toe.
• Horizontal distance B to F: $d = 105 - 65 = 40$ m.

Theoretical (Casagrande) discharge

Flow net

Base parabola ($x$ from F towards the upstream side): $x = \dfrac{y^2 - y_0^2}{2y_0}$

  A   B            crest 15 m
  ▽\__.\__________
   /   .·\          \
  /  entry ··..       \  2:1
 /________________F====\__ drain 30 m
   upstream toe     

Plot these points, correct the entry from A (perpendicular to the upstream slope), and end the top flow line at the drain at right angles. Draw 2 flow channels between the top flow line and the base, with equipotentials at equal head drops ($N_d \approx 7$); the head drop for each field is $25/7 = 3.6$ m.

Comparison

The flow net value depends on how accurately the fields are drawn, but with $N_f/N_d \approx 0.29$ it agrees with the theoretical value within a few percent.

In anisotropic soil the permeability differs in the two directions: $k_x$ (horizontal, usually larger) and $k_z$ (vertical). The flow net cannot be drawn directly with square fields, so the section is transformed.

Governing equation

Continuity for steady flow, $\dfrac{\partial v_x}{\partial x} + \dfrac{\partial v_z}{\partial z} = 0$, with Darcy's law $v_x = k_x\dfrac{\partial h}{\partial x}$ and $v_z = k_z\dfrac{\partial h}{\partial z}$ (sign neglected), gives

Transformation

Put $X = x\sqrt{\dfrac{k_z}{k_x}}$. Then $\dfrac{\partial}{\partial x} = \sqrt{\dfrac{k_z}{k_x}}\dfrac{\partial}{\partial X}$ and the equation becomes

which is the Laplace equation for an isotropic medium. So the section is redrawn with all horizontal dimensions multiplied by $\sqrt{k_z/k_x}$ (vertical dimensions unchanged), and the flow net with square fields is drawn on this transformed section.

Discharge

The flow across a vertical plane in the real section, through a vertical height $dz$:

This is Darcy's law in the transformed section with the equivalent permeability

For a flow net of square fields with total head loss $h$:

The same expression follows by considering the flow across a horizontal plane, so the result is independent of the direction.

Given

Upstream head 8 m, downstream head 1 m, so net head $H = 8 - 1 = 7$ m. $N_f = 9$, $N_d = 12$. $k = 3\times10^{-4}$ cm/s $= 3\times10^{-6}$ m/s.

Calculation

Answer: $q = 1.58\times10^{-5}$ m³/s per m ($\approx 1.36$ m³/day per m).

A layer of coarse material (a loaded filter) placed on top of the soil adds weight to resist the upward seepage force. The head loss in this material is negligible.

Given

Soil: $L = 1.15$ m, $n = 0.30$, $G = 2.7$. Upward seepage head at the base $h = 1.95$ m. Required factor of safety $F = 2$. Coarse layer of thickness $z$ with the same $G$ and $n$.

Submerged unit weight

Factor of safety against piping

At the base of the soil stratum, the effective stress with the load of the coarse layer:

Factor of safety = submerged weight / uplift:

For comparison, without the coarse layer $F = \dfrac{1.19\times1.15}{1.95} = 0.70 < 1$, so the stratum would pipe.

Answer: Thickness of coarse material required $\approx 2.13$ m.

Assumed figure

C = water surface, B = top of the soil, A = bottom of the soil. Depth of water above the soil $= H_2$ (C to B). Soil thickness $= H_1$ (B to A). Water flows upward through the soil with a total head loss $h$ between A and B. The saturated unit weight of soil is $\gamma_{sat}$ and $\gamma' = \gamma_{sat} - \gamma_w$.

Measure depths downward from C.

Stresses at the three points

Pore pressure at A is increased by $\gamma_w h$ above the hydrostatic value, since water enters at higher head. The excess head $h$ is lost uniformly through the soil (linear variation).

Diagrams

 depth   σ            u              σ'
 C  |    0            0              0
    |    ╲            ╲              
 B  |    γw·H2        γw·H2          0
    |      ╲            ╲              ╲
    |       ╲            ╲                ╲  (reduced
 A  |   γwH2+γsatH1  γw(H1+H2+h)   γ'H1 − γw·h   by seepage)

• Between C and B, $\sigma$ and $u$ increase together with slope $\gamma_w$ and $\sigma' = 0$.
• Between B and A, $\sigma$ increases linearly with slope $\gamma_{sat}$; $u$ increases linearly with slope $\gamma_w(1 + h/H_1) = \gamma_w(1+i)$; so $\sigma'$ increases linearly with slope $\gamma' - i\gamma_w$.
• When $i = h/H_1 = \gamma'/\gamma_w$, $\sigma'$ at A is zero: quick condition.