农场蓄水池：土方量、灌溉能力与防渗决策
A farmer wants to store rainwater in an on-farm pond. The excavation is rectangular at the top, 30 m long and 20 m wide and 2.5 m deep, with side slopes of 1:1 (horizontal to vertical) on all four sides. Local earthwork costs ₹150 per cubic metre of excavation. Runoff is expected to fill the pond once each rabi season, but seepage and evaporation are estimated to waste 30% of the stored volume before it can be used for irrigation. The stored water is to supplement irrigation of wheat at 6 cm depth per irrigation, where each additional irrigation is expected to add 12 q/ha at a wheat price of ₹2,400 per quintal. A contractor offers to line the pond with a geomembrane at ₹200 per square metre of wetted surface, which would cut losses from 30% to 8%; the membrane has a service life of about eight years.
1. Calculate the excavated volume of the pond and the total earthwork cost. Show the formula you use and the dimensions at the bottom and mid-depth.
Read the worked answer
With a 1:1 side slope and 2.5 m depth, every side moves in by 2.5 m horizontally.
- Top: 30 × 20 = 600 m²
- Bottom: (30 − 2×2.5) × (20 − 2×2.5) = 25 × 15 = 375 m²
- Mid-depth (1.25 m): (30 − 2×1.25) × (20 − 2×1.25) = 27.5 × 17.5 = 481.25 m²
Because the top and bottom rectangles are not similar, the prismoidal formula is the correct one (the frustum formula for similar bases would be wrong here):
V = (h/6) × (A_top + 4A_mid + A_bottom) V = (2.5/6) × (600 + 4×481.25 + 375) V = 0.4167 × (600 + 1,925 + 375) = 0.4167 × 2,900 = 1,208 m³ (approximately).
Cost: 1,208 × ₹150 = ₹1,81,250, i.e. about ₹1.81 lakh.
Sanity check: the average of top and bottom areas (487.5 m²) times depth gives 1,219 m³, within 1% of the prismoidal answer - so 1,208 m³ is reliable.
2. How much of the stored water is actually usable, and how many hectares of wheat can be given one 6 cm irrigation from one filling?
Read the worked answer
Usable water = stored volume × (1 − loss fraction) = 1,208 × 0.70 = 846 m³.
Water per hectare for one 6 cm irrigation = 0.06 m × 10,000 m²/ha = 600 m³/ha (this is the depth applied over the soil surface; application efficiency is already absorbed into the 30% loss figure, so it must not be deducted twice).
Area that can be irrigated once = 846 ÷ 600 = 1.41 ha.
If wheat needs two irrigations from this pond, the dependable area falls to about 0.7 ha; if only the critical crown-root and flowering stages are supported with one irrigation each on part of the field, the farmer should plan for roughly 1.4 ha-irrigations, not 1.4 ha of guaranteed irrigation. This distinction matters when deciding how much wheat area to commit.
3. Evaluate the lining offer using wetted surface area, the extra water saved, the value of that water and the payback period. Would you advise lining?
Read the worked answer
Wetted surface area:
- Bottom = 25 × 15 = 375 m²
- Slant length of each side = √(2.5² + 2.5²) = 3.54 m
- Two long sides = 2 × [(30+25)/2] × 3.54 = 194.5 m²
- Two short sides = 2 × [(20+15)/2] × 3.54 = 123.7 m²
- Total ≈ 693 m²
Lining cost = 693 × ₹200 = ₹1,38,600.
Extra water saved = 1,208 × (0.30 − 0.08) = 266 m³, which supports 266 ÷ 600 = 0.44 ha of one additional 6 cm irrigation.
Value of that water = 0.44 ha × 12 q/ha × ₹2,400/q = ₹12,764 per season.
Payback = 1,38,600 ÷ 12,764 ≈ 10.9 years, against an eight-year membrane life - the lining never repays itself within its service life.
Advice: do not line with the membrane. The pond is small and shallow, so the surface-to-volume ratio is unfavourable and the seepage loss is spread over a large area. Cheaper options come first: compact the bed and sides with a clay blanket, keep the pond silted bed sealed, reduce evaporation by minimising surface area (deeper and narrower rather than 30 × 20 m), and use the water on a high-value, water-critical irrigation rather than on a low-response one. The 12 q/ha response assumed for an additional irrigation is itself optimistic on a light soil, which makes the payback even longer.