Sizing a solar pump and storage for drip irrigation
Mr N. wants to irrigate 1.0 ha of vegetables with drip irrigation from a borehole. The peak crop water requirement is 6 mm/day. The drip system application efficiency is 90%, and he wants a 20% safety margin for wind and poor uniformity. The borehole yield is 6 m³/h and the static water level is 12 m below ground; the delivery tank is 8 m above ground and 30 m away horizontally. The pump must lift water to the tank, and the total pipe friction loss is estimated at 5 m. He plans to pump for 8 hours per day using a solar array and a 2.2 kW pump. A 10 m³ storage tank is available. He asks whether the system will meet the crop demand and whether the tank is large enough.
1. Calculate the peak daily water requirement and the required hourly flow over 8 pumping hours. Compare this with the borehole yield.
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Volume per day = 10,000 m² × 0.006 m = 60 m³. Gross requirement = 60/0.90 = 66.7 m³. Add 20% safety: 66.7 × 1.20 = 80 m³/day. Over 8 pumping hours the required flow is 80/8 = 10 m³/h. The borehole can supply only 6 m³/h, so in 8 hours it delivers 48 m³, well below the 80 m³ required. The system as planned cannot meet peak demand.
2. Calculate the total head and hydraulic power at 10 m³/h. Assess whether the 2.2 kW pump is adequate at 50% pump efficiency.
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Total head = static lift 12 m + delivery height 8 m + friction 5 m = 25 m. Hydraulic power = ρgQH = 1000 × 9.81 × (10/3600) × 25 ≈ 681 W. At 50% pump efficiency, shaft power ≈ 681/0.50 = 1.36 kW, so a 2.2 kW pump has enough power for 10 m³/h at 25 m. The limit is the borehole yield, not the motor. At 6 m³/h the hydraulic power is only about 409 W, so the pump would run well below capacity. The solar array must also be sized for the pump and for 8+ hours of useful sun, allowing for derating.
3. Evaluate the 10 m³ tank and propose practical changes to make the system reliable.
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A 10 m³ tank holds only 1 hour of flow at 10 m³/h, or 1.7 hours at 6 m³/h, so it is far too small for a day's demand of 80 m³. If pumping is direct, the tank is only a buffer. Options: (1) increase pumping hours to about 13.3 h at 6 m³/h to deliver 80 m³, which needs a larger tank and may run into cloudy weather; (2) reduce the irrigated area to about 0.6 ha (48 m³ pumped × 0.9/1.2 = 36 m³ net, and 36/60 = 0.6 ha); (3) add a second borehole or a larger storage tank. A tank of at least 80 m³ would give one day of autonomy, and 160 m³ two days. The 2.2 kW pump is not the main constraint; water supply and storage are.