Off-Grid Battery and Array Sizing for a Winter-Limited Cabin
A remote cabin uses 3.2 kWh/day of AC energy. The owner wants two days of autonomy without a generator and has chosen a 48 V lithium iron phosphate battery with 80% usable depth of discharge. The battery capacity is derated to 90% at the site’s 5°C winter design temperature. The inverter/charger efficiency is 92%, and the battery round-trip efficiency is 95%. The PV array will be mounted at a fixed tilt with a system derate factor of 0.75 for soiling, wiring, temperature, and age. Winter design insolation is 2.4 peak sun hours; summer is 5.2. The owner proposes a 1.2 kW array because the annual average is enough. The installer must size the battery bank in amp-hours and check whether the proposed array can recharge the bank in winter while serving the daily load. There is no grid and no backup generator.
1. Calculate the nominal battery capacity in amp-hours needed for two days of autonomy. Show each derating and state any assumption.
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Daily energy drawn from the battery by the inverter is 3.2 kWh / 0.92 = 3.48 kWh. For two days, that is 6.96 kWh.
If the 95% round-trip efficiency is applied conservatively to the discharge path as well, the required delivered energy becomes 6.96 / 0.95 = 7.33 kWh. The nominal battery energy must then be divided by both the 80% depth of discharge and the 90% temperature derate:
7.33 / (0.80 × 0.90) = 10.18 kWh
At 48 V, capacity = 10.18 kWh / 48 V = 212 Ah. A 220 Ah nominal bank is the practical choice. If the designer treats round-trip efficiency only on the charging side, the result is about 201 Ah, so 200 Ah is marginal and 220 Ah gives useful margin.
2. Size the PV array for winter and summer. Is the proposed 1.2 kW array adequate? What will happen in winter?
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The PV must supply the load through the inverter and battery. Required PV energy per day = 3.2 / (0.92 × 0.95 × 0.75) = 4.88 kWh. At the winter insolation of 2.4 peak sun hours, the array must be 4.88 / 2.4 = 2.03 kW. At the summer value of 5.2, the same calculation gives 0.94 kW.
The proposed 1.2 kW array in winter would deliver approximately 1.2 × 2.4 × 0.75 × 0.95 × 0.92 = 1.89 kWh AC per day, versus a 3.2 kWh load. The deficit is about 1.31 kWh/day, so the battery would be drawn down and the system would fail without a generator or load reduction. The annual average hides the winter failure; sizing must use the winter design month.
3. The owner wants to keep the 1.2 kW array and add a second battery instead. Evaluate that choice.
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A second battery does not fix an energy deficit; it only extends the time before the deficit empties the bank. With the 1.2 kW array, winter production is about 1.89 kWh/day against a 3.2 kWh/day load, so every day loses about 1.3 kWh. More storage would provide more autonomy days at the start of a cloudy period, but the system would still fail unless the array is enlarged, the load is reduced, the tilt or orientation is improved for winter, or a generator is added. The correct fix is to close the daily energy gap first, then size storage for the desired autonomy.