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Energy & Renewable Energy Engineering DISCUSSION
I want to run a small off-grid cabin: LED lighting, a laptop, a router and an efficient fridge, adding up to about 2.4 kWh per day, with the largest simultaneous load around 600 W. The site gets about 5 peak sun hours on an average day and closer to 3 in the worst month.
What is the calculation sequence from daily energy to the array power, the battery capacity at 24 V and the inverter rating? Sizing tools online give very different answers and I would like to understand the steps myself.
Start with the array. Required PV power = daily energy / (peak sun hours × system efficiency). The efficiency factor covers panel temperature, dirt, wiring, the charge controller, battery charge-discharge losses and the inverter; 0.7 to 0.75 is a reasonable planning figure for a battery system.
With 2.4 kWh, 5 sun hours and 0.75: 2400 / (5 × 0.75) = 640 W. That covers an average day only. Using the worst-month figure of 3 hours: 2400 / (3 × 0.75) = 1070 W. If the cabin is used all year, size for the worst month, or accept a generator or reduced loads in that season. Panels are usually the cheapest part of the system to oversize.
Battery capacity: Ah = daily energy × days of autonomy / (system voltage × usable depth of discharge × inverter efficiency). For one day of autonomy at 24 V with lead-acid batteries limited to 50 percent depth of discharge and a 90 percent efficient inverter: 2400 / (24 × 0.5 × 0.9) = 222 Ah. Two days needs about 445 Ah.
With lithium iron phosphate at 80 percent usable depth the same two days needs 2 × 2400 / (24 × 0.8 × 0.9) = 278 Ah. Choose autonomy from how many cloudy days in a row you must ride through. And check that the array can refill the bank: a battery far larger than the array can recharge in a day or two will sit partly discharged, which shortens the life of lead-acid batteries in particular.
Inverter: size it for power, not energy. The continuous rating should exceed the largest simultaneous load with margin, so for 600 W a 1000 W unit is sensible. Then check the surge rating against motor starts. A fridge compressor can draw several times its running power for a fraction of a second, and an inverter with too little surge capacity will trip even though the average load is small.
Also add the inverter's own idle consumption to the energy budget. A unit that draws 15 W with no load uses 0.36 kWh per day when left on, which is 15 percent of your whole daily load. For that reason do not pick an inverter far larger than you need.
Charge controller and wiring follow from the array. An MPPT controller's output current is roughly array power divided by battery voltage: 1070 W / 24 V = 45 A, so a 50 or 60 A controller, and its maximum input voltage must exceed the string's open-circuit voltage on the coldest morning. At 24 V the battery-to-inverter cable carries 600 / (24 × 0.9) = 28 A at full load and much more during a motor start, so it needs to be short and thick.
This is the reason to prefer 24 V or 48 V over 12 V for anything above a few hundred watts: the same power at 12 V doubles the current and quadruples the cable loss for a given cable.
Build the protection in from the start. Every battery can deliver thousands of amps into a short circuit, so put a fuse or DC-rated breaker sized for the cable as close to the battery positive as practical, and use DC-rated isolators on the PV side; AC breakers are not suitable for interrupting DC. Lead-acid batteries need ventilation because they release hydrogen when charging, and lithium batteries must have a battery management system and a charger profile that matches them.
The inverter output is ordinary mains voltage and just as dangerous. Earthing, residual-current protection and the AC wiring should follow your local electrical code, and in most places that part must be done or inspected by a licensed electrician.
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