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Energy & Renewable Energy Engineering DISCUSSION
I need to power a 300 W AC load from a 12 V 100 Ah battery and an inverter during outages. My simple calculation is 12 V × 100 Ah = 1200 Wh, so 4 hours, but a test with a lead-acid deep-cycle battery gave well under 2 hours before the inverter's low-voltage alarm sounded.
Where did the rest of the energy go, what is the right way to estimate runtime, and would a lithium iron phosphate battery of the same nominal capacity behave differently?
The 1200 Wh figure is the nominal energy, and three things reduce what you can use. First, the inverter: at about 90 percent efficiency, a 300 W load draws 333 W from the battery, which is 27.8 A at 12 V. Second, the rated capacity is measured at a slow discharge, usually over 20 hours (5 A for a 100 Ah battery); at higher currents a lead-acid battery delivers less. Third, you should not use all of it: regularly discharging lead-acid below about 50 percent shortens its life considerably.
On paper, even with no other effects, 1200 Wh × 0.9 / 300 W is 3.6 hours to completely flat, and 1.8 hours to 50 percent.
The rate effect is described by Peukert's relation: t = H × (C / (I × H))^k, where C is the rated capacity at the H-hour rate and k is the Peukert exponent, around 1.1 to 1.3 for lead-acid types. Taking k = 1.2 with C = 100 Ah, H = 20 h and I = 27.8 A: t = 20 × (100 / 556)^1.2 = 20 × 0.128 = 2.6 hours to fully discharged.
So at this current the battery behaves like a 71 Ah battery rather than 100 Ah. Stopping at 50 percent gives about 1.3 hours, which agrees with your test. The exponent varies with battery type and age, so treat the result as an estimate and use the manufacturer's capacity table for different discharge rates if there is one.
The inverter alarm also plays a part. Under a 28 A load the terminal voltage sags because of the battery's internal resistance and the cable resistance, so the inverter sees its low-voltage threshold earlier than the state of charge alone would suggest. Thin or long battery cables make it worse: 28 A through a total of 20 mΩ of cable and connections drops 0.56 V before the inverter even starts.
Keep the battery cables short and heavy, check every crimp and terminal, and measure the voltage at the battery posts and at the inverter terminals under load. A large difference between the two readings is lost runtime.
A lithium iron phosphate (LiFePO4) battery of 12.8 V and 100 Ah holds a nominal 1280 Wh and behaves differently in each of those respects. Its capacity is almost independent of discharge rate at these currents, its voltage stays nearly flat until it is close to empty, and using 80 to 90 percent of it routinely is normal. Estimated runtime: 1280 × 0.8 × 0.9 / 300 = about 3.1 hours, more than twice the lead-acid figure from the same nominal amp-hours.
It also wastes less energy in charging. Conditions apply: it needs its built-in battery management system, a charger with a lithium profile, and it must not be charged below freezing unless the battery is designed for that.
For sizing in the other direction, work from the energy you need: battery Wh = load power × hours / (inverter efficiency × usable fraction). Three hours of 300 W is 900 Wh at the load, so 900 / (0.9 × 0.5) = 2000 Wh of lead-acid, about 170 Ah at 12 V before the rate effect, or 900 / (0.9 × 0.8) = 1250 Wh of LiFePO4, about 100 Ah.
Whichever type you use, fit a fuse close to the battery positive terminal sized for the cable, since a 12 V battery can deliver a very large short-circuit current. Charge lead-acid batteries in a ventilated place, because they give off hydrogen, and do not improvise lithium packs from loose cells without a proper management system.
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