Soil Erosion and Watershed Management: USLE, Terraces and Bunds, Waterways, Gully Control, Spillways, Earthen Dams and Water Harvesting
2. Soil loss estimation: the Universal Soil Loss Equation
The Universal Soil Loss Equation predicts long-term average annual sheet and rill erosion: A = R · K · L · S · C · P, with A the soil loss (t/ha/year in metric use). The factors are defined relative to a unit plot — 22.13 m long on a uniform 9 per cent slope, kept in continuous clean-tilled fallow up and down the slope — for which L, S, C and P are all 1.
| Factor | What it represents | Range and behaviour |
|---|---|---|
| R | rainfall erosivity, the annual sum of storm EI₃₀ values | set by climate; larger for intense tropical rain |
| K | soil erodibility, loss per unit R on the unit plot | higher for silts, lower for clays and well-aggregated soils |
| L | slope-length factor | 1 at 22.13 m; rises with length |
| S | slope-steepness factor | 1 at 9 per cent; rises steeply with gradient |
| C | cover and management factor | 1 for bare fallow; far below 1 under dense cover |
| P | support-practice factor | 1 for up-and-down tillage; lower for contouring, strip cropping, terracing |
Worked: R = 500, K = 0.3, LS = 1.2, C = 0.4 and P = 0.5 give A = 500 × 0.3 × 1.2 × 0.4 × 0.5 = 36 t/ha/year. Because the equation is a product, it is also used backwards — to find the largest C (the crop system) or P (the practice) that keeps A below the tolerable soil loss: with R = 600, K = 0.25, LS = 1.5 and P = 0.6, holding A at 10 t/ha/year needs C ≤ 10/135 = 0.074. The USLE does not predict gully erosion or deposition, and single-storm estimates use its modified form (MUSLE) with a runoff term in place of R; RUSLE revises its factors.
3. Biological and engineering measures: terraces, bunds and vegetative waterways
Biological (agronomic) measures work by cover and roughness: contour farming, strip cropping (contour, field and buffer strips), mulching, cover crops, crop rotation, conservation tillage, agroforestry and grassed buffers. Engineering (mechanical) measures shorten the slope length and store or safely convey runoff: contour bunds (level earthen embankments along contours, for low-rainfall areas with permeable soils, where water is to be held), graded bunds (laid on a gentle longitudinal grade to drain excess water safely in higher-rainfall areas or less permeable soils), and terraces.
Bench terraces convert a steep slope into a flight of level or nearly level steps and are the standard treatment for cultivated hill slopes. Their geometry is fixed by the vertical interval VI (the drop between benches) and the land slope S per cent: the horizontal interval is HI = 100 VI/S, and with vertical risers the bench width equals HI; a sloping riser of U horizontal to 1 vertical takes U × VI of it. A 1 m vertical interval on a 20 per cent slope gives 5 m benches. With vertical risers and cut balanced against fill, each bench has a triangular cut and fill of height VI/2 over half its width, so the earthwork per metre length is (W/2)(VI/2)/2 = W·VI/8, and there are 10 000/W metres of bench per hectare: the earthwork is V = 1250 × VI m³/ha, independent of the bench width — 1500 m³/ha for a 1.2 m interval.
Vegetative (grassed) waterways carry runoff from terraces, graded bunds and field outlets down the slope without eroding their own bed. They are shaped parabolic (the most natural and stable form), trapezoidal or triangular, sized by the peak rate from the rational method, and designed with Manning’s equation for a permissible velocity that depends on the grass cover and soil — the grass is what allows a velocity that bare soil could not take. For a parabolic section of top width T and depth D, the area is A = (2/3)TD: a 6 m by 0.5 m waterway has A = 2 m² and at a design velocity of 1.25 m/s carries 2.5 m³/s.
4. Gully control structures, spillways and earthen dams
A gully is stabilised by stopping the water that feeds it at the head, reducing the grade of its bed and establishing vegetation. Temporary structures — brushwood, loose-rock and woven-wire check dams — trap sediment and slow the flow long enough for vegetation to take hold. Permanent structures are spillways that lower water safely through a fall in grade:
- Drop spillway: a weir (straight or box inlet) over which water falls vertically into a stilling basin that dissipates its energy; for drops of up to about 3 m where the gully bed is stable below.
- Drop-inlet (pipe) spillway: a vertical riser feeding a conduit through an embankment, so that water is stored upstream and released in a controlled way; as the head rises the flow passes from weir control at the riser crest to orifice control and finally to full pipe flow.
- Chute spillway: an open, steep, lined channel with an inlet, a conveyance section and an energy-dissipating outlet (such as the SAF stilling basin); for large drops and high flows.
Earthen dams for farm ponds and small reservoirs are homogeneous (one soil), zoned (an impervious core with pervious shells) or diaphragm type (a thin impervious membrane). They fail by overtopping (inadequate spillway or freeboard), seepage and piping (water finding a path through or under the dam), and sloughing or slope failure (steep or saturated slopes, rapid drawdown). Seepage is analysed through the phreatic line, which in a homogeneous dam with a horizontal filter is the Kozeny parabola: with h the water depth and d the horizontal distance from the filter face to where the parabola meets the reservoir (with Casagrande’s correction), its focal parameter is y₀ = √(h² + d²) − d, and the seepage per unit length is q = k·y₀. A 12 m head with d = 16 m gives y₀ = 20 − 16 = 4 m, and with k = 10⁻⁶ m/s, q = 4 × 10⁻⁶ m³/s per metre.
5. Watershed management: characterization, land capability, water budgeting, harvesting, check dams and farm ponds
A watershed is the area draining to a common outlet, and it is characterised by its size, shape, drainage network, slope, soils and land use. Shape indices compare it with a circle or square: the form factor R_f = A/L_b² (L_b the basin length), the circularity ratio (basin area over the area of a circle of the same perimeter) and the elongation ratio; a compact, circular basin gives a sharper, earlier peak than an elongated one. The drainage network is ordered by Strahler’s scheme — unbranched streams are order 1, two streams of order n join to make order n + 1 — and summarised by the bifurcation ratio R_b = N_u/N_(u+1) (typically 3 to 5) and the drainage density D_d = ΣL/A, the total stream length per unit area; high drainage density means impervious soils and a flashy response. A basin of 50 km² and 10 km length has a form factor of 0.5; 120 km of streams in 60 km² is a drainage density of 2 km/km².
Land-use capability classification groups land into eight classes by the severity of its limitations (slope, erosion hazard, depth, wetness, salinity, climate). Classes I to IV are suitable for cultivation, with limitations and the need for conservation practices increasing from I (few limitations) to IV (very severe, occasional cultivation only); classes V to VII are unsuited to cultivation and are kept under permanent vegetation — pasture, grazing and forestry; class VIII is fit only for wildlife, recreation and water supply. Subclasses name the kind of limitation (e for erosion, w for wetness, s for soil, c for climate).
Water budgeting closes the balance over a watershed for a period: P = Q + ET + ΔS + G (precipitation equals runoff, evapotranspiration, change in soil and surface storage, and deep percolation), and tells how much can be harvested without starving downstream users. Rainwater harvesting is in situ (contour furrows, ridges, micro-catchments that hold rain where it falls) or ex situ (collecting runoff into tanks, ponds and recharge structures); the harvestable volume is runoff coefficient × rainfall × catchment area — 0.2 × 0.6 m × 5 ha (50 000 m²) = 6000 m³. Check dams across drainage lines hold water, trap silt and recharge groundwater; farm ponds — dugout or embankment type — store runoff for life-saving irrigation, and are sized from the runoff volume, with allowances for seepage and evaporation losses and a spillway for the excess.
Key takeaways
- Water erosion is detachment plus transport; its types run splash, sheet, rill, gully and ravine as flow concentrates. Saltation dominates wind erosion.
- USLE A = RKLSCP on a 22.13 m, 9 per cent unit plot; being a product, it is solved backwards for the C or P that meets a tolerance.
- Bench terrace: HI = 100 VI/S; with vertical risers the earthwork is 1250 × VI m³/ha whatever the bench width.
- Drop, drop-inlet and chute spillways lower water through increasing drops; a drop inlet passes from weir to orifice to pipe flow.
- Seepage through a homogeneous dam with a horizontal filter is q = k·y₀, y₀ = √(h² + d²) − d.
- Land capability classes I–IV are arable, V–VII permanent vegetation, VIII wildlife and recreation; D_d = ΣL/A, R_f = A/L_b².
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
For a field, R = 500, K = 0.3, LS = 1.2, C = 0.4 and P = 0.5 (in consistent metric units). What is the average annual soil loss by the USLE, in t/ha/year?
Numerical answer — type the value.
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Answer: 36
A = R·K·LS·C·P = 500 × 0.3 × 1.2 × 0.4 × 0.5 = 36 t/ha/year. Leaving out P because "support practice" sounds optional gives 72 — the equation is a product, and every factor counts.A field has R = 600, K = 0.25, LS = 1.5 and P = 0.6. What is the largest cover-management factor C that keeps the soil loss at or below 10 t/ha/year (to three decimal places)?
Numerical answer — type the value.
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Answer: 0.074
C = A/(R·K·LS·P) = 10/(600 × 0.25 × 1.5 × 0.6) = 10/135 = 0.074. Forgetting P gives 10/225 = 0.044, a needlessly strict crop system; the contouring already credited in P lets a cover with C up to 0.074 do the job.Bench terraces with vertical risers are to be made on a land slope of 20 per cent with a vertical interval of 1 m. What is the width of each bench, in m?
Numerical answer — type the value.
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Answer: 5
HI = 100 VI/S = 100 × 1/20 = 5 m, and with vertical risers the whole horizontal interval is bench. Using VI × S = 20 m inverts the relation; a gentler slope needs wider benches, not narrower.Bench terraces with vertical risers and balanced cut and fill are built with a vertical interval of 1.2 m. What is the volume of earthwork per hectare, in m³?
Numerical answer — type the value.
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Answer: 1500
Cut per metre of bench = ½ × (W/2) × (VI/2) = W·VI/8; metres of bench per hectare = 10 000/W; so V = 10 000·VI/8 = 1250 × 1.2 = 1500 m³/ha, whatever W is. Counting cut and fill together doubles it to 3000, but the fill is the same soil moved, not extra earthwork.A watershed of 60 km² contains a total stream length of 120 km. What is its drainage density, in km/km²?
Numerical answer — type the value.
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Answer: 2
D_d = ΣL/A = 120/60 = 2 km/km². Inverting it (0.5 km²/km) gives the constant of channel maintenance, a different index.A basin has an area of 50 km² and a basin length of 10 km. What is its form factor?
Numerical answer — type the value.
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Answer: 0.5
R_f = A/L_b² = 50/10² = 0.5. Dividing by the length rather than its square (5 km) gives a width, not a dimensionless shape index.A fourth-order basin has 27 first-order, 9 second-order and 3 third-order streams and one fourth-order stream. What is its bifurcation ratio?
Numerical answer — type the value.
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Answer: 3
R_b = N_u/N_(u+1) = 27/9 = 9/3 = 3/1 = 3, constant across orders here. Dividing the total number of streams by the highest order (40/4 = 10) is not the bifurcation ratio.A farm pond is to collect runoff from a 5 ha catchment that receives 600 mm of seasonal rainfall with a runoff coefficient of 0.2. What volume of water can it harvest, in m³?
Numerical answer — type the value.
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Answer: 6000
V = C × P × A = 0.2 × 0.6 m × 50 000 m² = 6000 m³. Leaving the rainfall in millimetres gives six million, and forgetting the runoff coefficient gives the whole 30 000 m³ of rain, most of which infiltrates or evaporates.A homogeneous earthen dam with a horizontal toe filter holds a 12 m depth of water. The horizontal distance from the filter face to the corrected start of the phreatic line is 16 m, and the permeability is 1 × 10⁻⁶ m/s. What is the seepage per metre length of the dam, in units of 10⁻⁶ m³/s?
Numerical answer — type the value.
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Answer: 4
y₀ = √(h² + d²) − d = √(144 + 256) − 16 = 20 − 16 = 4 m, and q = k·y₀ = 10⁻⁶ × 4 = 4 × 10⁻⁶ m³/s per metre. Using q = kh (12 × 10⁻⁶) overstates it threefold, and √(h² + d²) alone (20 × 10⁻⁶) forgets to subtract d.A parabolic grassed waterway has a top width of 6 m and a flow depth of 0.5 m. At a design velocity of 1.25 m/s, what discharge does it carry, in m³/s?
Numerical answer — type the value.
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Answer: 2.5
A = (2/3)TD = (2/3) × 6 × 0.5 = 2 m², so Q = AV = 2 × 1.25 = 2.5 m³/s. Treating the section as a rectangle (6 × 0.5 = 3 m²) gives 3.75 m³/s, and as a triangle (1.5 m²) gives 1.875 m³/s.The "unit plot" on which the USLE factors are defined is:
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Answer: A — 22.13 m long on a uniform 9 per cent slope, in continuous clean-tilled fallow
On the unit plot — 22.13 m (72.6 ft), 9 per cent, bare tilled fallow up and down the slope — L, S, C and P are all 1, so its measured loss per unit R defines K. The other options change the length, slope or cover and so would not give unit values of those factors.Which of the following statements about land-use capability classification are correct?
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Answer: A — Classes I to IV are suitable for cultivation; B — Class VIII land is suited only to wildlife, recreation and water supply; C — Classes V to VII are best kept under permanent vegetation such as pasture or forest
Limitations increase from class I to VIII. Class I has few or no limitations and is the best arable land, so the fourth statement reverses the scale; the first three describe the arable, permanent-vegetation and non-agricultural groups correctly.In wind erosion, the mode of soil movement that carries the largest part of the load and drives the others is:
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Answer: A — saltation
Saltating grains hop along, pick up energy from the wind, and on landing both knock larger grains into creep and splash fine particles into suspension. Suspension carries dust farthest and is the most visible, which makes it the tempting wrong answer; sheet flow belongs to water erosion.As the head over the inlet of a drop-inlet (pipe) spillway rises, the control of flow passes in the order:
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Answer: A — weir flow, then orifice flow, then full pipe flow
At low heads water spills over the riser crest as over a weir; as the riser submerges, the inlet acts as an orifice; once the conduit runs full, pipe-flow friction and the total head control the discharge. The reversed order describes a falling, not a rising, head.