A 60 A Controller, 4S2P Array and a Battery Bank With One Dead Cell
A farmer's off-grid system runs household lights, fans and a small chiller: eight 250 Wp polycrystalline modules (Voc 37.8 V, Vmp 30.5 V, Isc 8.6 A, Imp 8.2 A) wired as two parallel strings of four in series, a charge controller marked '60 A MPPT, max PV input 150 V', and a 24 V bank of twelve 2 V, 400 Ah tubular cells (C10) commissioned 14 months ago. Nightly load, logged with a clamp meter and energy logger, is 3.6 kWh. The customer reports that lights dim by 9 pm, the inverter cuts out at 24.0 V, and the cells need watering every three weeks. Specific gravities after a full charge are 1.19, 1.18, 1.20, 1.17, 1.19, 1.18, 1.19, 1.20, 1.18, 1.19, 1.17 and 1.09. Winter minimum ambient at the site is about 5 °C. The farmer now wants to add a 1 hp (0.75 kW) AC pump running two hours a day and says 'just add two more panels'.
1. Check the array-to-controller match, including the cold-temperature Voc, and quantify what the 60 A rating is costing.
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Voltage check (cold)
- String Voc at 25 °C = 4 × 37.8 = 151.2 V, which is already above the controller's 150 V limit.
- At the 5 °C winter minimum, Voc rises by roughly 0.3 %/°C: 151.2 × (1 + 0.003 × 20) ≈ 160 V. The controller is being over-voltaged every cold morning; it is surviving on margin it does not have, and one cold snap can take out the input stage. This alone is an unsafe design.
Current check
- Array power ≈ 2 kWp; charging at ~27 V, that is 2000 / 27 ≈ 74 A.
- A 60 A controller can only pass about 60 × 27 ≈ 1.62 kW, so roughly 19 % of array output is clipped on every good day, and clipping is worst exactly when the battery most needs the charge (midday, after a deep night).
Correct specification
- Replace with an 80–100 A MPPT rated at 200 V PV input, or reconfigure to 2S4P (Voc 75.6 V, Vmp 61 V, input current 32.8 A) which is far friendlier to the controller and to cable sizing.
- Re-do the string voltage at the site's record low temperature, not at 25 °C. A controller must be chosen on cold Voc, never on Voc at STC.
2. Interpret the specific gravity readings and the watering interval. What has failed, what is the mechanism, and what do you do about it?
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What the numbers say. A fully charged tubular cell should read about 1.250–1.260. Ten cells at 1.17–1.20 are at roughly 50–70 % state of charge after a 'full' charge - chronic undercharging, not a load problem. The cell at 1.09 is effectively dead (sulphated or shorted).
The mechanism, in sequence
- The 60 A clipping and the over-voltage risk meant the bank rarely reached full charge, so lead sulphate hardened on the plates.
- The shorted cell drops 2 V out of the bank; the controller still regulates to a 28.8 V absorb setpoint, which is now 28.8 / 11 = 2.62 V per remaining cell instead of 2.40 V.
- That over-voltage on the healthy cells boils electrolyte - hence watering every three weeks and the falling specific gravities elsewhere. So the watering symptom is a consequence of the dead cell, not a separate fault. Do not simply tell the farmer to water more often.
Action
- Equalise the bank (about 2.55–2.60 V/cell for 2–3 h) and re-check specific gravities; if the low cell does not recover, replace it - at 14 months a single-cell swap is acceptable if the rest pass a discharge test, otherwise replace the bank.
- Correct settings after the fix: absorb ~28.8 V, float ~27.0 V, equalise ~31.0 V monthly, all temperature-compensated.
- Leave the low-voltage disconnect at 24.0 V under load (~2.0 V/cell); it is protecting the bank, not causing the dimming.
3. Assess the 1 hp pump request on energy, battery depth of discharge and starting surge, and state what you would honestly tell the farmer.
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Energy
- Pump: 0.75 kW × 2 h = 1.5 kWh AC, about 1.65 kWh at the battery after inverter losses.
- New daily load = 3.6 + 1.65 ≈ 5.25 kWh, i.e. 5.25 / 24 ≈ 219 Ah at 24 V.
- Against a 400 Ah bank that is 55 % depth of discharge every day - deep-cycle tubular cells will not survive that for long even when healthy, and this bank is already damaged.
- Array side: in December Punjab gives roughly 3.5–4 peak sun hours; 2 kWp × 3.8 × 0.7 (system losses) ≈ 5.3 kWh, before battery round-trip efficiency of about 85 %. The system would be running at a deficit in winter even with a perfect battery.
Surge. A 1 hp induction motor draws roughly 3× running current for a second or two - about 2.2 kW. The inverter must be rated for that surge and the battery must supply it; a small inverter will trip on start even when the energy is available.
What to tell him. 'Two more panels' is not wrong, it is just not enough: the honest package is two more modules, an 80–100 A MPPT with the right voltage window, and a bank of about 600 Ah (300 Ah usable at 50 % DoD) - or keep the pump on grid/diesel and use solar for the house loads. Show him the kWh arithmetic rather than arguing; farmers accept numbers about units faster than numbers about amps.