Sizing a Farm Pond and Testing a Paddy Plan
A watershed committee plans one farm pond on a 15 ha catchment and wants to irrigate paddy from it. Annual rainfall is 1,100 mm, of which about 85% falls between June and October; the block's runoff-plot estimate for this soil and land use gives a runoff coefficient of 0.30. The pit is planned at 50 m × 40 m at the base, 3.5 m deep, with 0.5 m of freeboard above top water level and 0.5 m of the depth left as dead storage for silt. Evaporation and seepage together are taken as 25% of live storage each year. Nine farmers, each holding 0.33 ha in the command, want to grow paddy on the 3 ha below the pond; paddy in this block needs about 1,000 mm of irrigation water per hectare. They expect the pond to supply the whole 3 ha from a single filling and have asked the committee to release the estimate for approval.
1. Using the given figures, test whether the pond can irrigate 3 ha of paddy. Show your calculation.
Read the worked answer
Catchment yield: 15 ha × 1.1 m × 0.30 = 4.95 ha·m = 49,500 m³. Runoff is not the limiting factor; the catchment could fill the pit several times over.
Storage: gross volume ≈ 50 × 40 × 3.5 = 7,000 m³ (assuming near-vertical sides; with 1:1 battering the true figure is lower and the estimate should be reworked from the section). Dead storage = 50 × 40 × 0.5 = 1,000 m³, leaving 6,000 m³ live. Losses at 25% = 1,500 m³, leaving about 4,500 m³ usable.
Demand: 3 ha × 1,000 mm = 3 × 10,000 = 30,000 m³.
Shortfall: 30,000 − 4,500 = 25,500 m³, so the pond meets about 15% of the plan. Paddy on 3 ha is not feasible; one hectare alone would need 10,000 m³, more than twice the usable storage. The estimate should not be approved on the paddy assumption, and conveyance losses make the real position worse.
2. Propose an irrigation and cropping plan that the 4,500 m³ can actually support, and justify it.
Read the worked answer
Convert the usable storage into a defensible command. Protective irrigation for upland crops in this zone is normally two to three applications of 5–6 cm at critical stages (establishment, flowering, pod or cob filling).
At 5 cm per irrigation, one hectare takes 500 m³ per round. Three rounds on 3 ha = 3 × 3 × 500 = 4,500 m³, which exactly matches usable storage - but leaves no cushion, so plan for 2.5 ha or reduce to two rounds on 3 ha. A safer recommendation is protective irrigation to about 2.5 ha of a short-duration upland crop such as green gram, cowpea or maize, with the remaining 0.5 ha rainfed.
Paddy should be dropped from the command, or confined to about 0.3–0.4 ha of the lowest land with seepage and return flow counted. Explain to the nine farmers that the pond is a life-saving device, not a canal: it turns a drought year from crop failure into a modest harvest, which is a real and bankable gain.
3. The committee wants the pond to last 20 years. What design and catchment treatments would you insist on before technical sanction?
Read the worked answer
Check the dead storage against the sediment load. With 15 ha of catchment losing roughly 5 t/ha/yr and a bulk density near 1.3, sediment is about 75 t/yr or 58 m³/yr. The 1,000 m³ dead storage therefore lasts roughly 17 years on paper, and less if the uplands stay bare - so the 20-year target is not met without catchment treatment.
Insist on: (i) vegetative filter strips and contour bunds on the erodible upper catchment before or alongside the excavation, since a pond below untreated land is a silt trap; (ii) two or three small check dams or brushwood barriers in the feeder channel to intercept bed load; (iii) an inlet that enters above top water level with a silt trap, not a direct cut; (iv) a defined spillway or waste weir, because a 0.5 m freeboard alone will not pass an abnormal storm; (v) fencing and a desilting provision in the annual maintenance; and (vi) a written maintenance sharing arrangement among the nine beneficiaries, since a pond with no desilting plan is a 10-year asset being financed as a 20-year one.