Water Resources HubWater topic 17 of 18 — free theory
Sedimentation & Erosion
How soil leaves a site, how sediment moves once it reaches the channel, and how engineers trap it — the USLE, settling basins, and channel protection.
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The USLE — predicting average annual soil loss
The Universal Soil Loss Equation is the workhorse for estimating sheet-and-rill erosion from a field or a construction site. It is empirical — fitted to thousands of plot-years of measured data — so treat it as a well-calibrated estimator, not a law of physics:
A = R · K · LS · C · P
| A | average annual soil loss, tons/acre/year (US customary units) |
| R | rainfall–runoff erosivity factor — climate-driven, roughly 50 in the arid west to 500+ in the Gulf states |
| K | soil erodibility factor — typically 0.02 (resistant clay) to 0.69 (highly erodible silt); most loams sit around 0.2–0.45 |
| LS | combined slope-length/steepness factor, dimensionless — grows quickly with steeper, longer slopes |
| C | cover-management factor, dimensionless — 1.0 for bare tilled soil down to ~0.001 under dense mulch or forest litter |
| P | support-practice factor, dimensionless — 1.0 with no practices, roughly 0.5–0.8 with contouring, strip-cropping, or terracing |
C and P are the designer’s levers — R, K, and LS are essentially fixed by the site. LS already folds slope length and steepness into one factor, so don’t multiply by a separate slope term.
How sediment moves — shear, Shields, and load types
Water starts moving a particle when the shear stress it applies to the bed exceeds the particle’s critical shear stress — the Shields concept. The applied boundary shear in a channel is:
τ = γ R S
| τ | average boundary shear stress on the bed |
| γ | unit weight of water (9.81 kN/m³ or 62.4 lb/ft³) |
| R | hydraulic radius |
| S | energy slope (≈ bed slope in uniform flow) |
Below critical shear nothing moves. Just above it, grains roll and hop along the bed (bed load); finer material lifts into the flow (suspended load); the very fine wash load passes straight through a reach. Finer grains have lower critical shear, so a flow that barely ripples gravel can carry silt in suspension.
Settling basins — the overflow rate controls everything
An ideal settling basin removes every particle whose settling velocity vs is at least as large as the overflow rate (surface loading rate):
vo = QAs td = VQ
| vo | overflow rate = design settling velocity captured |
| Q | inflow rate |
| As | basin surface area — not volume |
| td | detention time; V is the basin volume |
The classic exam slip is dividing by volume instead of surface area. Depth sets detention time, not removal — a deeper basin holds water longer but captures the same particle sizes.
Holding the channel together — permissible shear and riprap
Where velocities or shear exceed what the native bed tolerates, engineers armour the channel. The design check is the same Shields-style comparison: size the riprap (or choose the lining) so the critical shear of the protection exceeds the applied shear with margin. Isbash-type relations tie the required stone size to velocity squared — double the velocity and the stone weight needed roughly quadruples. Where the exam gives you a permissible-velocity or permissible-shear table, the workflow is: compute applied τ = γRS, compare, upsize protection until it passes.
PE depth: trap efficiency and sediment rating
A reservoir’s trap efficiency — the fraction of incoming sediment it keeps — rises with the capacity-to-inflow ratio (the Brune-curve idea): a large reservoir relative to its inflow traps nearly everything, while a small flood-control pool passes much of its sediment downstream.
Sediment rating: Qs = aQb (b usually 2–3)
Sediment load climbs far faster than discharge — doubling the flow roughly quadruples to octuples the load. That nonlinearity is why one big flood can deliver more sediment than a decade of ordinary flows, and why the design storm, not the average year, sizes most sediment structures.
PE trap: the same reservoir that tames floods (see Topic 18) is quietly filling with sediment and losing the storage its routing depends on. Always ask what the trap efficiency implies for long-term capacity.
Worked example USLE on a construction site
Given:
- R = 250, K = 0.32, LS = 1.4.
- With straw mulch and contouring: C = 0.10, P = 0.75.
- Without controls: C = 1.0, P = 1.0.
Solution:
- With controls: A = R·K·LS·C·P = 250×0.32×1.4×0.10×0.75. First 250×0.32 = 80; 80×1.4 = 112; 112×0.10 = 11.2; 11.2×0.75 = 8.4 tons/acre/yr.
- Without controls: A = 250×0.32×1.4×1.0×1.0 = 112 tons/acre/yr.
- The two management factors (C and P) cut predicted loss by a factor of 112/8.4 ≈ 13 — that is exactly what C and P are for.
Answer: ≈ 8.4 tons/acre/yr with mulch and contouring, versus ≈ 112 tons/acre/yr uncontrolled.