Sizing a Solar Pump When the Well, Not the Array, Is the Limit
A panchayat in a plateau block of Jharkhand wants a solar pump for its dug well. The well is 1.2 m in diameter, the static water level is 12 m below ground, and the safe drawdown is 3 m; dry-season recharge measured over two weeks is about 1,200 L/h. The pump must lift water a further 3 m and push it 400 m through a 50 mm pipe to a 5,000 L overhead tank that has to be filled once every day. Peak-sun hours in July are 4.5. Vendor A quotes a 3 kWp array with a 1.5 kW pump rated 3,000 L/h at 17 m total head, for ₹3.4 lakh. Vendor B quotes a 1 kWp array with a 0.5 kW pump rated 1,200 L/h at the same head, for ₹2.2 lakh. The panchayat argues for Vendor A because a bigger array 'will always give more water', and the village has had a pump run dry before.
1. Work out whether the tank can be filled from the well in a single sunny day, and state which quoted number actually limits the system. Show the arithmetic.
Read the worked answer
Daily demand is 5,000 L. Over about 4.5 usable sun hours the average flow needed is 5,000 ÷ 4.5 ≈ 1,110 L/h, say 1,200 L/h.
Well buffer: drawdown 3 m in a 1.2 m diameter well; area = π × 0.6² = 1.13 m²; stored volume = 1.13 × 3 = 3.4 m³ ≈ 3,400 L.
Vendor B: 1,200 L/h matches dry-season recharge, so there is no net drawdown; 4.5 h × 1,200 = 5,400 L, enough for the 5,000 L tank.
Vendor A: 3,000 L/h exceeds recharge by 1,800 L/h. The 3,400 L buffer is exhausted in about 3,400 ÷ 1,800 ≈ 1.9 h, after which the pump sucks air unless it is cycled or throttled.
The limiting number is the 1,200 L/h recharge, not the array size or the pump head. A bigger array cannot create water the aquifer does not deliver.
2. Estimate the hydraulic power needed to lift 1,200 L/h through the 17 m total head and compare it with the two pump ratings. What does this tell you about array sizing?
Read the worked answer
Hydraulic power = ρgQH. Q = 1,200 L/h = 1.2 ÷ 3,600 = 3.33 × 10⁻⁴ m³/s; H = 17 m.
P = 1,000 × 9.81 × 3.33 × 10⁻⁴ × 17 ≈ 56 W.
At a realistic wire-to-water efficiency of about 35%, electrical input is 56 ÷ 0.35 ≈ 160 W; even at 20% efficiency it is under 300 W. A 0.5 kW motor therefore has ample margin, while a 1.5 kW motor cannot be used at the well's safe yield and would only run in short bursts.
Daily energy: 5,000 L × 17 m = 85,000 kg·m ≈ 833 kJ ≈ 0.23 kWh of hydraulic work, roughly 0.7 kWh electrical. A 1 kWp array yields about 4.5 kWh on a good day, so the array is not the constraint either. The well is.
3. The panchayat insists on Vendor A. What would have to be added to make it safe, and what is the risk if nothing is added?
Read the worked answer
To run Vendor A safely you need: a well-level electrode or float that cuts the pump before water falls below the safe drawdown; a tank-level float that stops the pump when the tank is full; a non-return valve; and either a variable-frequency drive or a throttling valve plus a timer to hold average output near 1,200 L/h. With cycling, the well needs recovery time, so effective daily yield does not rise much.
Without these, the pump runs dry within about two hours, loses cooling and lubrication, and can burn out its motor; repeated over-pumping also silts and can destabilise a dug well.
The extra ₹1.2 lakh buys capacity the aquifer cannot supply. Put the savings into the storage tank, plumbing and level controls.