Sizing a Gully-Plug Check Dam Above a 3 ha Paddy
A 12 ha micro-watershed in a semi-arid district is to be treated with a gully-plug check dam. Soils are sandy loam with a curve number of 78 under AMC II. The 25-year, 24-hour rainfall is 140 mm. Mean annual sediment yield from the untreated catchment is 5 t/ha/year, trap efficiency of the planned structure is 85%, deposited bulk density is 1.3 t/m³, and the design life is 20 years. The proposed earth embankment is 2.5 m high with a storage capacity of 9,000 m³ at the emergency spillway crest. The longest flow path is 400 m at an average slope of 3%; the 25-year, 20-minute intensity is 90 mm/h, and the rational coefficient for the treated catchment is 0.35. The village expects the structure to protect 3 ha of paddy downstream and to supply one supplementary irrigation of 5 cm.
1. Compute the 25-year runoff volume at AMC II and at AMC III. Is the 9,000 m³ storage adequate?
Read the worked answer
AMC II: S = 25,400/78 − 254 = 71.6 mm; Ia = 0.2S = 14.3 mm.
Q = (140 − 14.3)²/(140 − 14.3 + 71.6) = 15,800/197.3 = 80.1 mm
Volume = 80.1 mm × 120,000 m² = 9,612 m³ - already above the 9,000 m³ provided, before any sediment.
AMC III: CN = 23(78)/(10 + 0.13 × 78) = 1,794/20.14 = 89.1; S = 31.2 mm, Ia = 6.2 mm.
Q = (140 − 6.2)²/(133.8 + 31.2) = 17,908/164.96 = 108.6 mm, i.e. 13,032 m³, about 36% more than AMC II.
The 25-year storm in this semi-arid setting is most likely to arrive on wet antecedent soil, so AMC III is the honest design case. Storage is inadequate by roughly 45% and the embankment would be overtopped unless the spillway carries the excess.
2. Allocate dead storage for sediment and re-check the live capacity against the irrigation demand.
Read the worked answer
Sediment: 12 ha × 5 t/ha/year = 60 t/year; trapped = 0.85 × 60 = 51 t/year; over 20 years = 1,020 t; volume = 1,020/1.3 = 785 m³.
Live storage = 9,000 − 785 = 8,215 m³, against a requirement near 9,600–13,000 m³.
Irrigation demand is small by comparison: 3 ha × 0.05 m = 1,500 m³, only 18% of live storage, so the irrigation function is not the binding constraint - flood attenuation is.
Judgement: do not simply raise the embankment. Either build two smaller structures in series, each sized to its own sub-catchment (which also spreads recharge and reduces spillway risk), or keep this site as a flood-attenuation structure with a properly sized spillway and drop the irrigation expectation to a single 5 cm application. Presenting the sediment allowance separately is essential; a design that hides it inside total capacity will fail in year 15, not year 20.
3. Estimate the peak discharge and recommend the spillway and freeboard arrangement.
Read the worked answer
Rational peak = CIA/360 = 0.35 × 90 × 12/360 = 1.05 m³/s.
Against a 2.5 m embankment, the real hazard is not the stored volume but the rate at which it arrives while the reservoir is already full from a previous storm. A vegetated emergency spillway is appropriate, but it must be sized with a margin - at least 1.5× the computed peak, i.e. about 1.6 m³/s - and cut in undisturbed ground, never in the fill.
Freeboard should be at least 0.5 m above the spillway crest, plus settlement allowance. Because the catchment is only 12 ha with a 400 m flow path, the time of concentration is short (of the order of 20–30 minutes), so the peak is flashy and warning time is minimal; the structure must pass the design storm without attendance.
Judgement: accept the site only if the spillway is sized for the AMC III volume and 1.5× the rational peak; otherwise move to the two-structure layout, which halves both the routed peak and the consequence of failure.