Sizing a 1.2 ha Tomato Drip Block Against a 6 L/s Borewell
Arun farms 1.2 ha of tomato near Jodhpur in a 120 m × 100 m block. Rows run the 100 m length and are 1.5 m apart, giving 80 rows. His borewell yields a tested 6 L/s and the submersible holds 2.5 bar at the filter outlet. Peak crop water use is 6 mm/day; the sandy loam takes water at about 12 mm/h before ponding. Canal water has EC 1.9 dS/m, and tomato yield is known to start declining above ECe 2.5 dS/m. The dealer who quoted the job selected 16 mm laterals with 4 L/h emitters every 40 cm, which is 10 L/h per metre of row, and proposed running the whole block at once. Nothing is installed yet. The farmer's two worries are whether he can irrigate every day through the peak period and whether 16 mm laterals will work at 100 m length.
1. Can the 6 L/s borewell run the whole 1.2 ha block at once? If not, divide the block into sets and give the set time and daily pumping hours. Show the arithmetic.
Read the worked answer
No. Full-block flow = 80 rows × (100 m × 10 L/h/m = 1,000 L/h = 1 m³/h) = 80 m³/h ≈ 22.2 L/s, about 3.7 times the well yield.
- Flow available per set = 6 L/s = 21.6 m³/h, so rows per set = 21.6 ÷ 1.0 = 21 rows (0.315 ha); the remainder gives a fourth, smaller set (17 rows).
- Application rate = 10 L/h per metre of row ÷ 1.5 m row spacing = 6.7 mm/h, comfortably below the 12 mm/h infiltration limit, so no runoff risk.
- Net depth per day = 6 mm; add about 15% leaching for EC 1.9 dS/m → ~7 mm. Set time = 7 ÷ 6.7 ≈ 1.05 h, round to 1 h.
- Four sets × 1 h = ~4.2 h/day of pumping, about 90 m³/day against a 72 m³ net requirement - consistent once leaching and efficiency are counted.
Practical consequence: the farmer needs a four-valve manifold at the head and a simple timer; four valve changes per day is the real operating cost, not water.
2. The quoted 16 mm lateral carries 1 m³/h over 100 m with non-compensating 4 L/h emitters at 1.0 bar. Check velocity and pressure variation and state what you would change.
Read the worked answer
Velocity check: 16 mm lateral has an internal diameter near 13.6 mm, area ≈ 1.45 × 10⁻⁴ m². Q = 1 m³/h = 2.78 × 10⁻⁴ m³/s, so v = 2.78 × 10⁻⁴ ÷ 1.45 × 10⁻⁴ ≈ 1.9 m/s - above the usual 1.5 m/s lateral limit and a scour/erosion risk.
Friction check: at 1 m³/h in a 13.6 mm pipe, Hazen-Williams (C = 140) gives a total friction loss of the order of 25–35 m over 100 m; even with a multi-outlet factor around 0.35–0.4 the loss is roughly 10 m (1 bar), far more than the ~0.4 bar that keeps non-compensating emitter flow within ±10–20%.
Change: go to 20 mm laterals (v ≈ 1.1 m/s, loss roughly 4 m over the run) and, because 4 m is still marginal on a 1-bar emitter, specify pressure-compensating emitters with a pressure regulator at each block inlet. Alternatively keep 16 mm but halve the run length to 50 m. Either way, install a gauge at the far end of the longest lateral so the claim can be checked on site, not argued on paper.
3. The farmer now wants to double the area to 2.4 ha using the same well. Give a reasoned recommendation, including what would have to change.
Read the worked answer
The water balance says yes on paper, the operation says not yet.
- Net demand at 2.4 ha × 6 mm = 144 m³/day; with ~15% leaching ≈ 165 m³/day. At 21.6 m³/h that is 7.6 h/day of pumping, which the well can physically deliver.
- But sets double to about eight per day, and the four-hour pumping window becomes a seven-to-eight-hour window with eight valve changes - much harder to manage with hired labour, and there is no reserve for a pump breakdown, filter cleaning or a power cut.
- Salinity is the second constraint: at EC 1.9 dS/m the leaching fraction must be maintained, so there is no water to waste on inefficiency.
Recommendation: do not expand while pumping direct to the field. Add storage (a 50 m³ tank filled overnight) and a booster, or raise well yield, so that irrigation is decoupled from pump run time. Expansion without storage simply converts a water problem into a reliability problem.