Six irrigations, a six-hour power window and a sloping field
A farmer in Fatehpur has 1.0 hectare of wheat sown on 22 November on a clay loam. He owns a 5 hp tube well delivering 600 litres per minute, but power is available only for 6 hours a day, and the pump uses 1.8 litres of diesel per hour when the grid fails. Soil tests give a field capacity of 28 per cent and a permanent wilting point of 12 per cent on a gravimetric basis, with a bulk density of 1.35 g/cm³. His neighbour, who owns 1.5 hectares, wants to share the pump on alternate days. Canal water will arrive only in February. He plans six irrigations and has already spent ₹9,000 on diesel this season; last year the same field gave 44 quintals per hectare with seven irrigations. The field slopes about 12 cm from north to south: the north-west corner stays waterlogged for two days after every irrigation, while the south-east corner dries out first.
1. Calculate the net irrigation depth needed at the crown-root stage and the pumping hours it will take. Show each step.
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Step 1 - available water capacity of the soil. Gravimetric difference between field capacity and wilting point = 28% − 12% = 16%, i.e. 0.16 kg/kg. Volumetric available water = 0.16 × bulk density 1.35 = 0.216, i.e. 21.6 per cent by volume. In other words, 21.6 mm of water is available per 100 mm of soil depth.
Step 2 - for the crown-root-stage root zone. At this stage the effective root zone is about 0–30 cm (300 mm). Available water = 0.216 × 300 mm ≈ 65 mm for the whole root zone.
Step 3 - depth to apply. The usual rule is to refill when about half the available water is used, so as not to stress the crop: 65 mm × 0.5 ≈ 32 mm, i.e. about 3.2 cm net.
Step 4 - volume and pumping time. Volume = 0.032 m × 10,000 m² = 320 m³ = 320,000 litres. At 600 litres per minute: 320,000 ÷ 600 = 533 minutes ≈ 9 hours of pumping, which is 1.5 days of his 6-hour power window.
Step 5 - gross depth. Allow for application losses with check-basin flooding at about 75 per cent efficiency: gross depth ≈ 3.2 ÷ 0.75 ≈ 4.3 cm, and the pumping time rises to about 12 hours, i.e. two days.
Conclusion: a 4 cm irrigation over 1 ha needs roughly 12 pumping hours - two days of his window, not one. Planning six irrigations means about 72 pumping hours (roughly 130 litres of diesel) across the season.
2. Which of the six irrigations should he protect, and which can he drop if power stays short? Where does the February canal water fit?
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Protect these first:
- Crown-root initiation (CRI), about 20–25 days after sowing - around 12–17 December. This is the single most yield-critical irrigation in wheat; missing it costs more than any other single watering.
- Flowering, around 75–85 days - late February to early March. Water stress here causes floret sterility and directly reduces grain number.
- Milk and dough stages - March. This is where grain weight is built, and it also protects against the hot, dry winds of late March.
Drop if short: the tillering irrigation (about 40–45 days) is the safest to skip, because the soil profile is usually still well charged from CRI and winter evaporation is low; the jointing irrigation can also be merged with flowering in an emergency. Never drop CRI to save a later one - that is a false economy.
Practical scheduling: with 6 hours a day and a neighbour sharing on alternate days, each irrigation needs two pumping days plus a day of rest. Plan backwards from CRI: he must be ready to start pumping around 10 December. Fix an IW:CPE schedule rather than guessing - with a pan factor of 0.9 and open-pan evaporation near 4 mm/day in December, a 4 cm irrigation is due roughly every 10–11 days; in March, at 6–7 mm/day, that shortens to 6–7 days.
February canal water: it arrives too late for CRI, so it cannot replace the first tube well irrigation. Use it for the fourth or fifth irrigation, and use the saved pumping hours for flowering. Irrigate in the early morning, avoid windy afternoons, and stop irrigating 10–12 days before harvest.
3. The field slopes 12 cm and one corner waterlogs. Explain how this wastes water, and rank the measures that would give him more crop per litre.
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How the slope wastes water. On a 12 cm fall across the field, water applied at the top runs off before it can infiltrate at the bottom, so the south-east corner is under-irrigated while the north-west corner collects the surplus and stands for two days. The under-irrigated corner loses yield, the waterlogged corner suffers nitrogen leaching, poor aeration and yellowing. A single average depth therefore does not meet either end of the field: the field as a whole needs more water to satisfy the dry corner, and the extra water damages the wet corner.
Ranked measures:
- Laser levelling or at least a careful land plane - removes the 12 cm fall and typically saves 15–25 per cent of irrigation water per season while raising yield. It is the highest-return single investment, and subsidy is often available.
- Check basins of the right size instead of free flooding - divide the field into small, level basins so each gets a measured 4 cm, cutting runoff and deep percolation.
- An underground pipeline for conveyance - an earthen channel can lose a large share of the pumped water before it reaches the field; a buried pipe delivers nearly all of it.
- Sprinkler irrigation - saves on the order of 30–40 per cent of water in wheat and suits a 6-hour power window, but the capital cost is high; check the subsidy before deciding.
- Pump and engine maintenance, and continuous sharing - a worn impeller or a high suction lift quietly reduces discharge below 600 litres per minute; running the pump in longer blocks with the neighbour avoids repeated starting losses.
Each of these multiplies the value of the same 6-hour window; buying more diesel without fixing the slope and the channel simply buys more runoff.