Permeability and Layered Soils – Lecture Notes
Permeability and Layered Soils – Lecture Notes
Focus of today: permeability, constant head tests (example solved), discharge velocity (Darcy velocity), seepage velocity, and how to handle layered soils (parallel vs. series flow).
Key event: Worked through an example of a constant head test on a soil sample to compute the coefficient of permeability and related velocities, then discussed multi-layer soils and how to combine layers into an equivalent permeability.
Quick recap: permeability tests
Permeability (k) characterizes how easily a fluid moves through a porous medium.
Constant head test vs. falling head test:
Constant head test: head difference Δh is maintained constant while measuring flow through a sample.
Falling head test: head difference decreases as water drains; used when permeability is low.
In this lecture, the emphasis is on constant head test for the solved example.
Example 3: Constant head test – given data
Sample height (length): L = 15\ \text{cm}
Cross-sectional area: A = 60\ \text{cm}^2
Water collected: Q_i = 40.5\ \text{cm}^3 over time t = 15\ \text{s}
Head difference: \Delta h = 24\ \text{cm}
Porosity of sand: n = 0.55
Step 1: Set up the main formula
Permeability (coefficient) from the constant head test:
k = \frac{Q L}{A \Delta h \ t}
Here, treat all lengths in cm so units are consistent.
Step 2: Plug in numbers
k = \frac{(40.5\ \text{cm}^3)(15\ \text{cm})}{(60\ \text{cm}^2)(24\ \text{cm})(15\ \text{s})} = 2.8\times 10^{-3}\text{ cm/s} = 2.8\times 10^{-4}\text{ m/s}
Note: The calculation yields k = 0.0028\ \text{cm/s} = 2.8\times 10^{-4}\ \text{m/s} (conversion to m/s shown).
Step 3: Darcy (discharge) velocity and seepage velocity
Hydraulic gradient, per unit length, is given by:
i = \frac{\Delta h}{L} = \frac{24}{15} = 1.6
Discharge velocity (Darcy velocity):
v = k \cdot i = (2.8\times 10^{-4}\ \text{m/s})(1.6) \approx 4.5\times 10^{-4}\ \text{m/s}
Seepage velocity (through voids):
v_s = \frac{v}{n} = \frac{4.5\times 10^{-4}}{0.55} \approx 8.1\times 10^{-4}\ \text{m/s}
Summary of results for Example 3:
Coefficient of permeability: k \approx 2.8\times 10^{-4}\ \text{m/s}
Discharge velocity (Darcy velocity): v \approx 4.5\times 10^{-4}\ \text{m/s}
Seepage velocity: v_s \approx 8.1\times 10^{-4}\ \text{m/s}
Important nuance: the head difference Δh is measured across the length of the sample in this setup. If Δh is measured between two points not spanning the full length, you must use the distance between those points for i.
Layered soils: multiple layers, 1D flow
Real soils are often stratified (multiple layers) with different properties. We need a way to characterize the overall flow through the layered system.
Two primary flow configurations in layered soils:
Flow parallel to the layers (horizontal seepage)
Flow perpendicular to the layers (vertical seepage)
Use a unit thickness (into the page) to simplify, so the cross-sectional areas scale with layer thicknesses.
1) Horizontal seepage (flow parallel to layers)
Basic idea: equivalent horizontal permeability, k_h^{eq}, is obtained by averaging layer permeabilities weighted by their thicknesses.
Derivation (conceptual): Q into layers equals Q into equivalent layer, with a common hydraulic gradient i. For unit thickness, area for layer j is Aj = hj.
Formula (standard):
kh^{eq} = \frac{\sum{j=1}^n hj kj}{\sum{j=1}^n hj} = \frac{\sumj hj kj}{H{tot}}
where H{tot} = \sumj h_j is the total thickness.
Boundary conditions: equal hydraulic gradient across all layers (one-dimensional, same L and Δh across layers).
Important intuition: for horizontal seepage, the most permeable layer often dominates the flow, similar to having a larger pipe in parallel branches.
2) Vertical seepage (flow perpendicular to layers)
Basic idea: equivalent vertical permeability, k_v^{eq}, is found from a series combination rule.
Formula (standard):
\frac{H{tot}}{kv^{eq}} = \sum{j=1}^n \frac{hj}{kj} = \sumj \frac{Hj}{kj}
Or equivalently, kv^{eq} = \frac{H{tot}}{\sumj \left( \frac{hj}{k_j} \right)}
Boundary conditions: Q is the same through each layer (conservation of mass) and the sum of head losses across layers equals the total head loss across the equivalent profile.
Important intuition: for vertical seepage, the least permeable layer controls the overall seepage (like the smallest pipe constraining the flow in series).
Key comparisons:
Horizontal flow (parallel layers): the overall permeability is controlled by the weighted average toward the more permeable layers; the flow tends to be governed by the most permeable layer in the sense of contributing more to the overall conductivity.
Vertical flow (perpendicular layers): the overall permeability is controlled by the least permeable layer, as the flow must pass through it in series.
Worked examples (layered soil) – two cases
Example 1: Horizontal seepage through a layered soil in a canal (1D, horizontal flow)
Problem setup: seepage takes place laterally through the side of a canal; 3 layers with thicknesses and permeabilities:
Layer 1: thickness h1 = 1.0\ \text{m},\; k1 = 2.3\times 10^{-7}\ \text{m/s}
Layer 2: thickness h2 = 1.5\ \text{m},\; k2 = 5.2\times 10^{-8}\ \text{m/s}
Layer 3: thickness h3 = 0.5\ \text{m},\; k3 = 2\times 10^{-8}\ \text{m/s}
Total thickness: H = h1 + h2 + h_3 = 3.0\ \text{m}
Apply the horizontal formula:
kh^{eq} = \frac{\sumj hj kj}{H} = \frac{h1 k1 + h2 k2 + h3 k3}{H}
Compute numerator: h1 k1 + h2 k2 + h3 k3 = (1.0)(2.3\times 10^{-7}) + (1.5)(5.2\times 10^{-8}) + (0.5)(2\times 10^{-8})
Numerically: = 2.3\times 10^{-7} + 7.8\times 10^{-8} + 1.0\times 10^{-8} = 3.18\times 10^{-7}
Therefore: k_h^{eq} = \frac{3.18\times 10^{-7}}{3.0} = 1.06\times 10^{-7}\ \text{m/s} (approximately)
Observation: the equivalent permeability is close to the most permeable layer (Layer 1, k_1 = 2.3\times 10^{-7}\ \text{m/s}).
Takeaway: the flow is dominated by the most permeable layer in a parallel/ horizontal arrangement.
Example 2: Vertical seepage through layered soil (flow perpendicular to layers)
Problem setup: upward flow through three layers with vertical permeabilities:
Layer 1: height h1 = 1.5\ \text{m},\; k{v1} = 2\times 10^{-8}\ \text{m/s}
Layer 2: height h2 = 1.2\ \text{m},\; k{v2} = 0.3\times 10^{-6}\ \text{m/s} = 3\times 10^{-7}\ \text{m/s}
Layer 3: height h3 = 3.0\ \text{m},\; k{v3} = 0.8\times 10^{-5}\ \text{m/s} = 8\times 10^{-6}\ \text{m/s}
Total height: H{tot} = h1 + h2 + h3 = 5.7\ \text{m}
Apply the vertical formula:
\frac{H{tot}}{kv^{eq}} = \sum{j=1}^n \frac{hj}{k_{vj}}
Compute each term:
\frac{h1}{k{v1}} = \frac{1.5}{2\times 10^{-8}} = 7.5\times 10^{7}
\frac{h2}{k{v2}} = \frac{1.2}{3\times 10^{-7}} = 4\times 10^{6}
\frac{h3}{k{v3}} = \frac{3.0}{8\times 10^{-6}} = 3.75\times 10^{5}
Sum: ≈ 7.5\times 10^{7} + 4\times 10^{6} + 3.75\times 10^{5} \,\approx\ 7.54\times 10^{7}
Therefore: kv^{eq} = \frac{H{tot}}{\sum (hj/k{vj})} = \frac{5.7}{7.54\times 10^{7}} \approx 7.6\times 10^{-8}\ \text{m/s}
Observation: even though Layer 3 is relatively permeable, the equivalent vertical permeability is dominated by the least permeable layer (on the order of 10^{-8} m/s).
Takeaway: in vertical (series) flow, the least permeable layer controls seepage behavior.
Practical connections and reasoning
When splitting flow across layered soils:
Horizontal arrangement acts like parallel resistors: the overall permeability is a weighted average towards more permeable layers.
Vertical arrangement acts like series resistors: the overall permeability is controlled by the least permeable layer (largest sum of hj/kj).
Real-world relevance: boreholes, canals, and embankments often encounter layered soils; predicting seepage rates helps in designing cutoffs, drains, and waterproofing.
Important caveats:
These derivations assume steady, one-dimensional seepage and full saturation.
For two-dimensional seepage, the problem becomes more complex and requires different formulations.
Summary of key formulas to memorize
Constant head test permeability:
k = \frac{Q L}{A \Delta h \ t}
In the example, with Q = 40.5\ \text{cm}^3, L=15\ \text{cm}, A=60\ \text{cm}^2, \Delta h=24\ \text{cm}, t=15\ \text{s}, you get k = 0.0028\ \text{cm/s} = 2.8\times 10^{-4}\ \text{m/s}.
Darcy (discharge) velocity:
v = k \cdot i,\quad i = \frac{\Delta h}{L}
Seepage velocity:
v_s = \frac{v}{n}
Horizontal equivalent permeability (flow parallel to layers):
kh^{eq} = \frac{\sumj hj kj}{\sumj hj} = \frac{\sumj hj kj}{H{tot}}
Vertical equivalent permeability (flow perpendicular to layers):
\frac{H{tot}}{kv^{eq}} = \sumj \frac{hj}{k{vj}} \quad\Rightarrow\quad kv^{eq} = \frac{H{tot}}{\sumj \left(\frac{hj}{k{vj}}\right)}
Boundary conditions (conceptual):
Horizontal: equal hydraulic gradient across layers; total discharge is distributed by layer conductivities.
Vertical: equal discharge through layers; head losses add up to the total head loss.
Important qualitative takeaway:
Horizontal seepage is dominated by the most permeable layer (in a practical sense, the layer with the largest contribution to the sum).
Vertical seepage is dominated by the least permeable layer (the bottleneck controlling the series flow).
Connection to prior topics and practical implications
Built on Darcy’s law and the concept of hydraulic gradient from earlier lectures.
Connects lab-measurement concepts (permeability tests) with field applications (layered soils, seepage paths).
Provides a framework for simplifying complex stratified subsurface conditions into a single equivalent permeability for design and analysis.
If you have questions on any step, we can walk through the arithmetic again or apply these formulas to a new layered-soil scenario to reinforce the method.