Drip System Retrofit on a 1.5 ha Tomato Block: Well Yield, Block Operation and Salinity
A 1.5 ha tomato block (150 m × 100 m) in Nueva Ecija is laid out with rows 1.2 m apart running the 150 m length. The cooperator has already bought a 50 mm diameter PE mainline and 16 mm laterals with inline emitters spaced 0.30 m, each rated 1.6 L/h at 10 m head. Design peak crop evapotranspiration is 7.0 mm/day and the assumed drip irrigation efficiency is 90%. The only water source is a shallow tube well yielding 5 L/s that can be pumped for at most 10 hours per day. The farmer intends to run the whole field as one block starting at 6 a.m. Well water tests 1.5 dS/m; the crop is known to lose yield once the soil saturation extract exceeds about 2.5 dS/m. Two weeks after commissioning, plants at the far ends of the laterals are visibly smaller, and a gauge at the pump reads 1.4 bar while a gauge at the lateral inlet reads 0.6 bar.
1. Compute the total emitter discharge, the peak daily water requirement, and the daily pumping time. Can the field be run as a single block on this well? Show the hydraulics, including a velocity check on the 50 mm mainline.
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Emitter count. Laterals run the 150 m length, so each carries 150 / 0.30 = 500 emitters. Number of laterals = 100 m / 1.2 m = 83. Total = 41,500 emitters.
Block discharge. 41,500 × 1.6 L/h = 66,400 L/h = 66.4 m³/h = 18.4 L/s.
Demand. Gross depth = 7.0 / 0.90 = 7.78 mm/day. Volume = 0.00778 m × 15,000 m² = 116.7 m³/day. At the full 66.4 m³/h this is only 1.76 h of running, which looks easy but is misleading.
Well constraint. The well gives 5 L/s = 18 m³/h, so the field must be split into 66.4 / 18 = 3.7, i.e. 4 blocks of 16.6 m³/h (4.6 L/s each). Pumping time = 116.7 / 18 = 6.5 h, and with three change-overs the day fits inside the 10 h window.
Mainline velocity. v = Q/A. At full flow: (66.4/3600) / (π × 0.025²) = 0.01844 / 0.001963 = 9.4 m/s. Even at block flow: 0.00461 / 0.001963 = 2.35 m/s. Both exceed the 1.0–1.5 m/s range used for buried PE. A 50 mm main cannot serve even one block without excessive friction; the pump will run far back on its curve and deliver much less than 66 m³/h, which is consistent with the 0.8 bar loss already observed.
Conclusion. Single-block operation is hydraulically impossible. Operate four blocks and replace the main with 75 mm (1.05 m/s at block flow), or 90 mm if the run is long.
2. The soil test shows the field is beginning to salinize. Compute a leaching requirement for the well water and revise the pumping schedule. State the assumptions and the risks of applying the result blindly.
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Formula. Use the steady-state leaching requirement LR = ECw / (5·ECe − ECw), where ECw is the irrigation water salinity and ECe is the target saturation-extract salinity at the crop threshold.
Numbers. ECw = 1.5 dS/m, ECe = 2.5 dS/m, so LR = 1.5 / (12.5 − 1.5) = 1.5 / 11 = 0.13, about 13%.
Revised schedule. Leaching volume = 116.7 × 0.13 = 15.2 m³/day, for a total of 131.9 m³/day. At 18 m³/h that is 7.3 h of pumping plus change-overs, still inside the 10 h window but with much less slack.
Assumptions to state. The formula assumes uniform water application, steady state, no shallow water table, and that the crop is at its threshold. It is a long-term average, not a per-irrigation figure - the 13% is normally delivered by lengthening each set rather than by adding a separate leaching event.
Risks and refinements. (1) If the soil is already above threshold, set the target ECe higher (say 1.5 × threshold) and leach more. (2) Leaching only works if there is internal drainage; on a heavy clay with a shallow water table the salt will not leave the root zone and a drainage outlet or a change of crop is needed. (3) The salts go somewhere - check the receiving drain or groundwater. (4) Re-test soil ECe at the start and end of the season rather than relying on a single calculation, and note that tomato yield falls roughly 9–10% per dS/m above the 2.5 dS/m threshold.
3. Explain the 0.8 bar difference between the pump gauge and the lateral inlet gauge, and the poor growth at the lateral tails. Give a sequence of field checks and the corrective actions, in order of priority.
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Interpretation. A 0.8 bar (about 8 m) loss between the pump and the lateral inlet is far too large for a well-designed mainline; with a 50 mm main it is expected, because the velocity is above 2 m/s even at block flow and friction head grows with v². The low reading at the lateral inlet means the emitters are operating well below their rated 10 m head (1.6 L/h), so every plant is under-irrigated - and the far ends of the laterals, which are furthest along the friction path, receive the least. Hence the stunted tail plants rather than a uniform crop.
Check sequence. (1) Pressure at filter inlet, filter outlet, main inlet and lateral inlet - locate where the loss occurs. (2) Clean or replace the filter and check that it is not bypassing. (3) Look for leaks, broken connectors and open ends at the head of the field. (4) Dig up and inspect emitters at head, middle and tail; measure discharge into a can for one minute at several points to get the actual flow, not the catalog value. (5) Confirm the pump is not cavitating and that the suction line and impeller are clear.
Corrective actions, priority order. Replace or upsize the main to 75 mm; convert to pressure-compensating emitters or fit a pressure regulator at each block inlet; flush all laterals and treat with chlorine or acid if biological or mineral clogging is found; then re-commission with a pressure and flow log per block so uniformity can be verified before the next season.