Firstly let me say that this is with a view to using a lead-acid battery for backup purposes in case of power outage (mains power, a few lightbulbs, but with daily outages lasting a few hours)
So let's provide a data sheet for the battery, deep-cycle flooded lead-acid type:
For the given battery, the SPRE 12 225, the open-circuit cell voltage is:
- 100% charged 2.122V per cell
- 50% charged 2.017V per cell
- 20% charged 1.943V per cell (11.66V total)
- 10% charged 1.918V per cell (11.51V total)
Meanwhile the capacity of the battery is defined based on discharging (at constant current) in 10, 20, 48, 72, or 100 hours until the cell voltage falls to 1.75V (10.5V total).
According to the manufacturer's data, the total lifetime capacity of the battery (number of cycles * depth of discharge) doesn't appear to be significantly different between 20% and 80% DoD.
So if we want to know how much real capacity the battery has, it seems we can use 12V * C * 80%. The C10 figure is 179Ah, C20 is 204Ah and C100 is 225 Ah. So at C10, 17.9A, that's 12 & 179 * 0.8 = 1718.4W.
However, is it possible that the voltage drop due to the current & internal resistance of the battery is distorting the figures? Or is this insignificant?
The manufacturer do not quote below a C10 figure (17.9A, equivalent to around 170W after inverter & distribution losses). Is this the maximum current draw? Or could you go to, say, C5?
I notice from the manufacturer's datasheet for their renewal energy storage range
that the ratio of, say C10 to C100 isn't completely consistent across battery chemistry, however they do quote C5 numbers for some of the AGM & gel batteries. Is there in fact a possibility to use a higher discharge current for gel/AGM vs lead-acid? And as such perhaps a better battery chemistry for the deep cycle, fast discharge use case (let's say we we get a 1 hour powercut and want to discharge at, say C2, and have a generator for backup in the case that it goes over 1 hour)?
Secondly in terms of charging the battery, the SPRE 12 225 states 'a maximum charge current of 13% of C20 rate'. Does this mean 13% of the C20 capacity of 204Ah = 26.5 A charging current, and hence we can charge from 20% to 80% (122.4Ah), say, in just over 4.5 hours?